Page 1 Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 0 ? INTRODUCTION In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles. ? HISTORICAL FACTS EUCLID was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle. Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language. A king once asked Euclid, "Isn't there an easier way to understand geometry" Euclid replied : "There is no royalroad way to geometry. Every one has to think for himself when studying." THALES (640546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the wellknown and very useful theorem credited after his name : Thales Theorem. ? CONGRUENT FIGURES Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent figures are alike in every respect. Ex. 1 . Two squares of the same length. 2 . Two circle of the same radii. 3 . Two rectangles of the same dimensions. 4 . Two wings of a fan. 5 . Two equilateral triangles of same length. ? SIMILAR FIGURES Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. E uclid : Father of G eometry (about 300 B.C . G reece) T hales (640546 B.C .) SIMILAR TRIANGLES SIMILAR TRIANGLES Page 2 Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 0 ? INTRODUCTION In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles. ? HISTORICAL FACTS EUCLID was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle. Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language. A king once asked Euclid, "Isn't there an easier way to understand geometry" Euclid replied : "There is no royalroad way to geometry. Every one has to think for himself when studying." THALES (640546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the wellknown and very useful theorem credited after his name : Thales Theorem. ? CONGRUENT FIGURES Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent figures are alike in every respect. Ex. 1 . Two squares of the same length. 2 . Two circle of the same radii. 3 . Two rectangles of the same dimensions. 4 . Two wings of a fan. 5 . Two equilateral triangles of same length. ? SIMILAR FIGURES Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. E uclid : Father of G eometry (about 300 B.C . G reece) T hales (640546 B.C .) SIMILAR TRIANGLES SIMILAR TRIANGLES Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 Ex. 1. Any two line segments are similar. 2. Any two equilateral triangles are similar 3. Any two squares are similar. 4. Any two circles are similar. We use the symbol '~' to indicate similarity of figures. ? SIMILAR TRIANGLES ?ABC and ?DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional. i.e., when ?A = ?D, ?B = ?E, ?C = ?F and AB BC AC DE EF DF ? ? . A B C D E F And, we write ?ABC ~ ?DEF. The sign '~' is read as 'is similar to'. THEOREM1 (Thales Theorem or Basic Proportionality Theorem) : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Given : A ?ABC in which line ? parallel to BC (DE ?BC) intersecting AB at D and AC at E. To prove : AD AE DB EC ? Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ? AB and through D draw DG ? ?AC. A B C F G D E ? Proof : STATEMENT REASON 1. Area of ( ?ADE) = 1 2 (AD × EF) Area of ? = 1 base altitude 2 ? Area of ( ?BDE) = 1 2 (BD × EF) Page 3 Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 0 ? INTRODUCTION In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles. ? HISTORICAL FACTS EUCLID was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle. Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language. A king once asked Euclid, "Isn't there an easier way to understand geometry" Euclid replied : "There is no royalroad way to geometry. Every one has to think for himself when studying." THALES (640546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the wellknown and very useful theorem credited after his name : Thales Theorem. ? CONGRUENT FIGURES Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent figures are alike in every respect. Ex. 1 . Two squares of the same length. 2 . Two circle of the same radii. 3 . Two rectangles of the same dimensions. 4 . Two wings of a fan. 5 . Two equilateral triangles of same length. ? SIMILAR FIGURES Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. E uclid : Father of G eometry (about 300 B.C . G reece) T hales (640546 B.C .) SIMILAR TRIANGLES SIMILAR TRIANGLES Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 Ex. 1. Any two line segments are similar. 2. Any two equilateral triangles are similar 3. Any two squares are similar. 4. Any two circles are similar. We use the symbol '~' to indicate similarity of figures. ? SIMILAR TRIANGLES ?ABC and ?DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional. i.e., when ?A = ?D, ?B = ?E, ?C = ?F and AB BC AC DE EF DF ? ? . A B C D E F And, we write ?ABC ~ ?DEF. The sign '~' is read as 'is similar to'. THEOREM1 (Thales Theorem or Basic Proportionality Theorem) : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Given : A ?ABC in which line ? parallel to BC (DE ?BC) intersecting AB at D and AC at E. To prove : AD AE DB EC ? Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ? AB and through D draw DG ? ?AC. A B C F G D E ? Proof : STATEMENT REASON 1. Area of ( ?ADE) = 1 2 (AD × EF) Area of ? = 1 base altitude 2 ? Area of ( ?BDE) = 1 2 (BD × EF) Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 2. 1 AD EF Area ( ADE) AD 2 1 Area ( BDE) DB BD EF 2 ? ? ? ? ? ? By 1. 3. 1 AE DG Area ( ADE) AE 2 1 Area ( CDE) EC EC DG 2 ? ? ? ? ? ? Similarly 4. Area ( ?BDE) = Area ( ?CDE) ?s BDE and CDE are on the same base BC and between the same parallel lines DE and BC. 5. Area ( ADE) Area ( BDE) ? ? = AE EC By 3. & 4. 6. AD DB = AE EC By 1. & 5. Hence proved. THEOREM2 (Converse of Basic Proportionality Theorem) : If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. Given : A ?ABC and DE is a line meeting AB and AC at D and E respectively such that AD DB = AE EC To prove : DE ?BC B C D A E F Proof : STATEMENT REASON 1. If possible, let DE be not parallel to BC. Then, draw DF ?BC 2. AD DB = AF FC By Basic Proportionality Theorem. 3. AD DB = AE EC Given 4. ? AF FC = AE EC From 2 and 3. ? AF FC +1 = AE EC + 1 Adding 1 on both sides. ? AF FC AE EC FC EC ? ? ? By addition. ? AC AC FC EC ? AF + FC = AC and AE + EC = AC. ? FC = EC ? E and F coincide. But, DF ?BC. Hence DE ?BC. Hence, proved. Page 4 Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 0 ? INTRODUCTION In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles. ? HISTORICAL FACTS EUCLID was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle. Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language. A king once asked Euclid, "Isn't there an easier way to understand geometry" Euclid replied : "There is no royalroad way to geometry. Every one has to think for himself when studying." THALES (640546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the wellknown and very useful theorem credited after his name : Thales Theorem. ? CONGRUENT FIGURES Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent figures are alike in every respect. Ex. 1 . Two squares of the same length. 2 . Two circle of the same radii. 3 . Two rectangles of the same dimensions. 4 . Two wings of a fan. 5 . Two equilateral triangles of same length. ? SIMILAR FIGURES Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. E uclid : Father of G eometry (about 300 B.C . G reece) T hales (640546 B.C .) SIMILAR TRIANGLES SIMILAR TRIANGLES Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 Ex. 1. Any two line segments are similar. 2. Any two equilateral triangles are similar 3. Any two squares are similar. 4. Any two circles are similar. We use the symbol '~' to indicate similarity of figures. ? SIMILAR TRIANGLES ?ABC and ?DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional. i.e., when ?A = ?D, ?B = ?E, ?C = ?F and AB BC AC DE EF DF ? ? . A B C D E F And, we write ?ABC ~ ?DEF. The sign '~' is read as 'is similar to'. THEOREM1 (Thales Theorem or Basic Proportionality Theorem) : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Given : A ?ABC in which line ? parallel to BC (DE ?BC) intersecting AB at D and AC at E. To prove : AD AE DB EC ? Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ? AB and through D draw DG ? ?AC. A B C F G D E ? Proof : STATEMENT REASON 1. Area of ( ?ADE) = 1 2 (AD × EF) Area of ? = 1 base altitude 2 ? Area of ( ?BDE) = 1 2 (BD × EF) Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 2. 1 AD EF Area ( ADE) AD 2 1 Area ( BDE) DB BD EF 2 ? ? ? ? ? ? By 1. 3. 1 AE DG Area ( ADE) AE 2 1 Area ( CDE) EC EC DG 2 ? ? ? ? ? ? Similarly 4. Area ( ?BDE) = Area ( ?CDE) ?s BDE and CDE are on the same base BC and between the same parallel lines DE and BC. 5. Area ( ADE) Area ( BDE) ? ? = AE EC By 3. & 4. 6. AD DB = AE EC By 1. & 5. Hence proved. THEOREM2 (Converse of Basic Proportionality Theorem) : If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. Given : A ?ABC and DE is a line meeting AB and AC at D and E respectively such that AD DB = AE EC To prove : DE ?BC B C D A E F Proof : STATEMENT REASON 1. If possible, let DE be not parallel to BC. Then, draw DF ?BC 2. AD DB = AF FC By Basic Proportionality Theorem. 3. AD DB = AE EC Given 4. ? AF FC = AE EC From 2 and 3. ? AF FC +1 = AE EC + 1 Adding 1 on both sides. ? AF FC AE EC FC EC ? ? ? By addition. ? AC AC FC EC ? AF + FC = AC and AE + EC = AC. ? FC = EC ? E and F coincide. But, DF ?BC. Hence DE ?BC. Hence, proved. Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 4 Ex.1 In the adjoining figure, DE ?BC. ( i ) If AD = 3.4 cm, AB = 8.5 cm and AC = 13.5 cm, find AE. ( i i ) If AD 3 = DB 5 and AC = 9.6 cm, find AE. B C D A E Sol. ( i ) Since DE ?BC, we have AD AE AB AC ? ? 3.4 AE 8.5 13.5 ? ? 3.4 13.5 8.5 ? = 5.4 Hence, AE = 5.4 cm. ( i i ) Since DE ?BC, we have AD AE DB EC ? ? AE 3 EC 5 ? AD 3 (Given) DB 5 ? ? ? ? ? ? ? ? Let AE = x cm. Then, EC = (AC – AE) = (9.6 – x) cm. ? x 3 9.6 – x 5 ? ? 5x = 3(9.6 – x) ? 5x = 28.8 – 3x ? 8x = 28.8 ? x = 3.6. ? AE = 3.6 cm. Ex.2 In the adjoining figure, AD = 5.6 cm, AB = 8.4 cm, AE = 3.8 cm and AC = 5.7 cm. Show that DE ?BC. Sol. We have, AD = 5.6 cm, DB = (AB – AD) = (8.4 – 5.6) cm = 2.8 cm. AE = 3.8 cm, EC = (AC – AE) = (5.7 – 3.8) cm = 1.9 cm. ? AD 5.6 2 DB 2.8 1 ? ? and AE 3.8 2 EC 1.9 1 ? ? A D B E C Thus, AD AE DB EC ? ? DE divides AB and AC proportionally. Hence, DE ?BC Ex.3 In fig, PS PT = SQ TR and ?PST = ?PRQ. Prove that PQR is an isosceles triangle. [NCERT] Sol. It is given that PS PT = SQ TR So, ST ?QR [Theorem] P T R Q S Therefore, ?PST = ?PQR [Corresponding angles]  (1) Also, it is given that ?PST = ?PRQ ( 2 ) So, ?PRQ = ?PQR [From 1 and 2] Therefore PQ = PR [Sides opposite the equal angles] i . e . , PQR is an isosceles triangle. Ex.4 Prove that any line parallel to parallel sides of a trapezium divides the nonparallel sides proportionally (i.e., in the same ratio). OR ABCD is a trapezium with DC ?AB. E and F are points on AD and BC respectively such that EF ?AB. Show Page 5 Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 0 ? INTRODUCTION In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles. ? HISTORICAL FACTS EUCLID was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle. Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language. A king once asked Euclid, "Isn't there an easier way to understand geometry" Euclid replied : "There is no royalroad way to geometry. Every one has to think for himself when studying." THALES (640546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the wellknown and very useful theorem credited after his name : Thales Theorem. ? CONGRUENT FIGURES Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent figures are alike in every respect. Ex. 1 . Two squares of the same length. 2 . Two circle of the same radii. 3 . Two rectangles of the same dimensions. 4 . Two wings of a fan. 5 . Two equilateral triangles of same length. ? SIMILAR FIGURES Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. E uclid : Father of G eometry (about 300 B.C . G reece) T hales (640546 B.C .) SIMILAR TRIANGLES SIMILAR TRIANGLES Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 Ex. 1. Any two line segments are similar. 2. Any two equilateral triangles are similar 3. Any two squares are similar. 4. Any two circles are similar. We use the symbol '~' to indicate similarity of figures. ? SIMILAR TRIANGLES ?ABC and ?DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional. i.e., when ?A = ?D, ?B = ?E, ?C = ?F and AB BC AC DE EF DF ? ? . A B C D E F And, we write ?ABC ~ ?DEF. The sign '~' is read as 'is similar to'. THEOREM1 (Thales Theorem or Basic Proportionality Theorem) : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Given : A ?ABC in which line ? parallel to BC (DE ?BC) intersecting AB at D and AC at E. To prove : AD AE DB EC ? Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ? AB and through D draw DG ? ?AC. A B C F G D E ? Proof : STATEMENT REASON 1. Area of ( ?ADE) = 1 2 (AD × EF) Area of ? = 1 base altitude 2 ? Area of ( ?BDE) = 1 2 (BD × EF) Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 2 2. 1 AD EF Area ( ADE) AD 2 1 Area ( BDE) DB BD EF 2 ? ? ? ? ? ? By 1. 3. 1 AE DG Area ( ADE) AE 2 1 Area ( CDE) EC EC DG 2 ? ? ? ? ? ? Similarly 4. Area ( ?BDE) = Area ( ?CDE) ?s BDE and CDE are on the same base BC and between the same parallel lines DE and BC. 5. Area ( ADE) Area ( BDE) ? ? = AE EC By 3. & 4. 6. AD DB = AE EC By 1. & 5. Hence proved. THEOREM2 (Converse of Basic Proportionality Theorem) : If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. Given : A ?ABC and DE is a line meeting AB and AC at D and E respectively such that AD DB = AE EC To prove : DE ?BC B C D A E F Proof : STATEMENT REASON 1. If possible, let DE be not parallel to BC. Then, draw DF ?BC 2. AD DB = AF FC By Basic Proportionality Theorem. 3. AD DB = AE EC Given 4. ? AF FC = AE EC From 2 and 3. ? AF FC +1 = AE EC + 1 Adding 1 on both sides. ? AF FC AE EC FC EC ? ? ? By addition. ? AC AC FC EC ? AF + FC = AC and AE + EC = AC. ? FC = EC ? E and F coincide. But, DF ?BC. Hence DE ?BC. Hence, proved. Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 4 Ex.1 In the adjoining figure, DE ?BC. ( i ) If AD = 3.4 cm, AB = 8.5 cm and AC = 13.5 cm, find AE. ( i i ) If AD 3 = DB 5 and AC = 9.6 cm, find AE. B C D A E Sol. ( i ) Since DE ?BC, we have AD AE AB AC ? ? 3.4 AE 8.5 13.5 ? ? 3.4 13.5 8.5 ? = 5.4 Hence, AE = 5.4 cm. ( i i ) Since DE ?BC, we have AD AE DB EC ? ? AE 3 EC 5 ? AD 3 (Given) DB 5 ? ? ? ? ? ? ? ? Let AE = x cm. Then, EC = (AC – AE) = (9.6 – x) cm. ? x 3 9.6 – x 5 ? ? 5x = 3(9.6 – x) ? 5x = 28.8 – 3x ? 8x = 28.8 ? x = 3.6. ? AE = 3.6 cm. Ex.2 In the adjoining figure, AD = 5.6 cm, AB = 8.4 cm, AE = 3.8 cm and AC = 5.7 cm. Show that DE ?BC. Sol. We have, AD = 5.6 cm, DB = (AB – AD) = (8.4 – 5.6) cm = 2.8 cm. AE = 3.8 cm, EC = (AC – AE) = (5.7 – 3.8) cm = 1.9 cm. ? AD 5.6 2 DB 2.8 1 ? ? and AE 3.8 2 EC 1.9 1 ? ? A D B E C Thus, AD AE DB EC ? ? DE divides AB and AC proportionally. Hence, DE ?BC Ex.3 In fig, PS PT = SQ TR and ?PST = ?PRQ. Prove that PQR is an isosceles triangle. [NCERT] Sol. It is given that PS PT = SQ TR So, ST ?QR [Theorem] P T R Q S Therefore, ?PST = ?PQR [Corresponding angles]  (1) Also, it is given that ?PST = ?PRQ ( 2 ) So, ?PRQ = ?PQR [From 1 and 2] Therefore PQ = PR [Sides opposite the equal angles] i . e . , PQR is an isosceles triangle. Ex.4 Prove that any line parallel to parallel sides of a trapezium divides the nonparallel sides proportionally (i.e., in the same ratio). OR ABCD is a trapezium with DC ?AB. E and F are points on AD and BC respectively such that EF ?AB. Show Smilar Triangles – Aadhard OYCRP 97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi  01132044009 4 that AE BF = ED FC [NCERT] Sol. We are given trapezium ABCD. CD ?BA EF ?AB and CD both We join AC. It mets EF at O. In ?ACD, OE ?CD ? AO OC = AE ED . . . ( i ) E D C F A B O (Basic Proportionality Theorem) In ?CAB, OF ?AB ? CO CF OA FB ? [B.P.T] ? AO BF OC FC ? . . . ( i i ) From (i) and (ii) AE BF ED FC ? Hence, proved. Ex.5 Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. (Internal Angle Bisector Theorem) Sol. Given : A ?ABC in which AD is the internal bisector of ?A. 1 2 A C D B 3 4 E To Prove : BD AB DC AC ? Construction : Draw CE ?DA, meeting BA produced at E. Proof : STATEMENT REASON 1. ?1 = ?2 AD is the bisector of ?A 2. ?2 = ?3 Alt. ?s are equal, as CE ?DA and AC is the transversal 3. ?1 = ?4 Corres. ?s are equal, as CE ?DA and BE is the transversal 4. ?3 = ?4 From 1, 2 and 3. 5. AE = AC Sides opposite to equal angles are equal 6. In ?BCE, DA ?CE ? BD DC = BA AE By B.P.T. ? BD DC = AB AC Using 5 Hence, Proved.Read More
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