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Abc is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersect AC at D. Show that 1) D is the mid -point of ac 2) MD is perpendicular AC 3) CM=MA= ½AB ?
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Abc is a triangle right angled at C. A line through the mid-point M of...
Statement: ABC is a right-angled triangle with right angle at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.
To prove:
1) D is the midpoint of AC.
2) MD is perpendicular to AC.
3) CM = MA = 1/2 * AB.
Proof:
1) D is the midpoint of AC:
To prove that D is the midpoint of AC, we need to show that AD = DC.
Let's consider triangle ABC and its midpoint M. Since M is the midpoint of AB, we have MA = MB.
Now, since MD is parallel to BC, we can apply the midpoint theorem. According to the midpoint theorem, a line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
Therefore, we have MD || BC and MD = 1/2 * BC.
Since ABC is a right-angled triangle, we can use the Pythagorean theorem to relate the lengths of its sides. According to the Pythagorean theorem, AC^2 = AB^2 + BC^2.
Substituting AB = 2MA and BC = 2MD, we get AC^2 = (2MA)^2 + (2MD)^2.
Simplifying, we have AC^2 = 4MA^2 + 4MD^2.
Now, let's consider triangle ADC. According to the Pythagorean theorem, AD^2 = AC^2 + CD^2.
Substituting AC^2 = 4MA^2 + 4MD^2, we get AD^2 = 4MA^2 + 4MD^2 + CD^2.
Since MD is parallel to BC, angle AMD = angle C. Similarly, angle ADM = angle A.
Using the AA similarity criterion, we can conclude that triangle AMD ~ triangle ADC.
Therefore, the corresponding sides are proportional:
AD/AM = AC/AD.
Simplifying, we have AD^2 = AC^2.
Substituting AC^2 = 4MA^2 + 4MD^2, we get AD^2 = 4MA^2 + 4MD^2.
Now, comparing this equation with AD^2 = 4MA^2 + CD^2, we can conclude that CD^2 = 4MD^2.
Taking the square root of both sides, we get CD = 2MD.
Since AD = 2AM and CD = 2MD, we can conclude that AD = CD, which proves that D is the midpoint of AC.
2) MD is perpendicular to AC:
To prove that MD is perpendicular to AC, we need to show that angle MDC = 90 degrees.
Since MD is parallel to BC, we have angle MDC = angle C.
Since ABC is a right-angled triangle, we know that angle C is 90 degrees.
Therefore, angle MDC = 90 degrees, which proves that MD is perpendicular to AC.
3) CM = MA = 1/2 * AB:
Since D is the midpoint of AC, we have AD = DC.
Since MD is parallel to BC, we have MD = 1/2 * BC.
Substituting BC = 2MD
To prove:
1) D is the midpoint of AC.
2) MD is perpendicular to AC.
3) CM = MA = 1/2 * AB.
Proof:
1) D is the midpoint of AC:
To prove that D is the midpoint of AC, we need to show that AD = DC.
Let's consider triangle ABC and its midpoint M. Since M is the midpoint of AB, we have MA = MB.
Now, since MD is parallel to BC, we can apply the midpoint theorem. According to the midpoint theorem, a line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
Therefore, we have MD || BC and MD = 1/2 * BC.
Since ABC is a right-angled triangle, we can use the Pythagorean theorem to relate the lengths of its sides. According to the Pythagorean theorem, AC^2 = AB^2 + BC^2.
Substituting AB = 2MA and BC = 2MD, we get AC^2 = (2MA)^2 + (2MD)^2.
Simplifying, we have AC^2 = 4MA^2 + 4MD^2.
Now, let's consider triangle ADC. According to the Pythagorean theorem, AD^2 = AC^2 + CD^2.
Substituting AC^2 = 4MA^2 + 4MD^2, we get AD^2 = 4MA^2 + 4MD^2 + CD^2.
Since MD is parallel to BC, angle AMD = angle C. Similarly, angle ADM = angle A.
Using the AA similarity criterion, we can conclude that triangle AMD ~ triangle ADC.
Therefore, the corresponding sides are proportional:
AD/AM = AC/AD.
Simplifying, we have AD^2 = AC^2.
Substituting AC^2 = 4MA^2 + 4MD^2, we get AD^2 = 4MA^2 + 4MD^2.
Now, comparing this equation with AD^2 = 4MA^2 + CD^2, we can conclude that CD^2 = 4MD^2.
Taking the square root of both sides, we get CD = 2MD.
Since AD = 2AM and CD = 2MD, we can conclude that AD = CD, which proves that D is the midpoint of AC.
2) MD is perpendicular to AC:
To prove that MD is perpendicular to AC, we need to show that angle MDC = 90 degrees.
Since MD is parallel to BC, we have angle MDC = angle C.
Since ABC is a right-angled triangle, we know that angle C is 90 degrees.
Therefore, angle MDC = 90 degrees, which proves that MD is perpendicular to AC.
3) CM = MA = 1/2 * AB:
Since D is the midpoint of AC, we have AD = DC.
Since MD is parallel to BC, we have MD = 1/2 * BC.
Substituting BC = 2MD
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Abc is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersect AC at D. Show that 1) D is the mid -point of ac 2) MD is perpendicular AC 3) CM=MA= ½AB ?
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Abc is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersect AC at D. Show that 1) D is the mid -point of ac 2) MD is perpendicular AC 3) CM=MA= ½AB ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Abc is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersect AC at D. Show that 1) D is the mid -point of ac 2) MD is perpendicular AC 3) CM=MA= ½AB ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Abc is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersect AC at D. Show that 1) D is the mid -point of ac 2) MD is perpendicular AC 3) CM=MA= ½AB ?.
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