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If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse?
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If a perpendicular is drawn from the vertex containing the right angle...
Introduction:
In a right triangle, if a perpendicular is drawn from the vertex containing the right angle to the hypotenuse, then the two triangles formed on either side of the perpendicular are similar to each other and to the original triangle. Additionally, the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.
Proof of Triangle Similarity:
Let's consider a right triangle ABC, where angle B is the right angle. Draw a perpendicular from vertex B to the hypotenuse AC, and let the point of intersection be D.
Triangle ABD:
In triangle ABD, angle ABD is 90 degrees (as BD is perpendicular to AC). Therefore, angle ADB = 180 - 90 - A = 90 - A.
In triangle ABC, angle ABC = 90 degrees (given), so angle BAC = 180 - 90 - B = 90 - B.
Since angle ADB = angle BAC and angle ABD = angle ABC, by the Angle-Angle (AA) similarity criterion, triangle ABD is similar to triangle ABC.
Triangle CBD:
In triangle CBD, angle CBD is 90 degrees (as BD is perpendicular to AC). Therefore, angle CDB = 180 - 90 - C = 90 - C.
In triangle ABC, angle BCA = 90 degrees (given), so angle BAC = 180 - 90 - A = 90 - A.
Since angle CDB = angle BAC and angle CBD = angle BCA, by the Angle-Angle (AA) similarity criterion, triangle CBD is similar to triangle ABC.
Triangle ABD ~ Triangle CBD:
Since triangle ABD is similar to triangle ABC and triangle CBD is similar to triangle ABC, by the Transitive property of similarity, triangle ABD is similar to triangle CBD.
Proof of Perpendicular Square and Hypotenuse Product:
Let the lengths of the two parts of the hypotenuse AC be x and y, where x + y = AC.
Using the similarity of triangle ABD and triangle CBD, we can establish the following ratios:
AB/AD = BD/BC
AB/(AC - AB) = (AC - AB)/BC
AB/(x + y - AB) = (x + y - AB)/BC
Cross-multiplying, we get:
AB * BC = (x + y - AB)^2
AB * BC = (x + y)^2 - 2AB(x + y) + AB^2
Using the Pythagorean theorem, we know that AB^2 + BC^2 = AC^2, so AB * BC = x^2 + y^2
Therefore, x^2 + y^2 = (x + y)^2 - 2AB(x + y) + AB^2
Simplifying the equation, we get:
x^2 + y^2 = x^2 + 2xy + y^2 - 2AB(x + y) + AB^2
0 = 2xy - 2AB(x + y) + AB^2
0 = 2xy - 2AB(x + y) + AB^2
0 = 2xy - 2(x + y)(AB) + AB^2
0 =
In a right triangle, if a perpendicular is drawn from the vertex containing the right angle to the hypotenuse, then the two triangles formed on either side of the perpendicular are similar to each other and to the original triangle. Additionally, the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.
Proof of Triangle Similarity:
Let's consider a right triangle ABC, where angle B is the right angle. Draw a perpendicular from vertex B to the hypotenuse AC, and let the point of intersection be D.
Triangle ABD:
In triangle ABD, angle ABD is 90 degrees (as BD is perpendicular to AC). Therefore, angle ADB = 180 - 90 - A = 90 - A.
In triangle ABC, angle ABC = 90 degrees (given), so angle BAC = 180 - 90 - B = 90 - B.
Since angle ADB = angle BAC and angle ABD = angle ABC, by the Angle-Angle (AA) similarity criterion, triangle ABD is similar to triangle ABC.
Triangle CBD:
In triangle CBD, angle CBD is 90 degrees (as BD is perpendicular to AC). Therefore, angle CDB = 180 - 90 - C = 90 - C.
In triangle ABC, angle BCA = 90 degrees (given), so angle BAC = 180 - 90 - A = 90 - A.
Since angle CDB = angle BAC and angle CBD = angle BCA, by the Angle-Angle (AA) similarity criterion, triangle CBD is similar to triangle ABC.
Triangle ABD ~ Triangle CBD:
Since triangle ABD is similar to triangle ABC and triangle CBD is similar to triangle ABC, by the Transitive property of similarity, triangle ABD is similar to triangle CBD.
Proof of Perpendicular Square and Hypotenuse Product:
Let the lengths of the two parts of the hypotenuse AC be x and y, where x + y = AC.
Using the similarity of triangle ABD and triangle CBD, we can establish the following ratios:
AB/AD = BD/BC
AB/(AC - AB) = (AC - AB)/BC
AB/(x + y - AB) = (x + y - AB)/BC
Cross-multiplying, we get:
AB * BC = (x + y - AB)^2
AB * BC = (x + y)^2 - 2AB(x + y) + AB^2
Using the Pythagorean theorem, we know that AB^2 + BC^2 = AC^2, so AB * BC = x^2 + y^2
Therefore, x^2 + y^2 = (x + y)^2 - 2AB(x + y) + AB^2
Simplifying the equation, we get:
x^2 + y^2 = x^2 + 2xy + y^2 - 2AB(x + y) + AB^2
0 = 2xy - 2AB(x + y) + AB^2
0 = 2xy - 2AB(x + y) + AB^2
0 = 2xy - 2(x + y)(AB) + AB^2
0 =
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If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse?
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If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse?.
If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to the each other and to the original triangle also prove that the square of perpendicular is equal to the product of the length of the two part of the hypotenuse?.
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