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Finding Ratio of Sides in Triangle When Angles Are Given

In a triangle...

Question

In a triangle ABC, angle A=90°, CA=AB and D is a point on AB produced. Prove that

DC² - BD²= 2AB × AD

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Solution

In a triangle ABC,∠A = 90°, CA=AB and D is a point on AB produced.
According to Pythagoras theorem, DC² = AD² + AC²

DC² - BD² = (AD² + AC²) - (AD - AB)²
= (AD² + AC²) - (AD² + AB²- 2 AD . AB)
= AD² + AC² - AD² - AB²+ 2 AD . AB
= AC² - AB²+ 2 AD . AB
= AC² - AC²+ 2 AD . AB [Since AC = AB]
= 2 AD . AB

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Similar questions

A B C

A = 90^{o}, C A = A B

D

A B

{D C}^{2} - {B D}^{2} = 2 A B . A D

△ A B C

∠ A = 90^{o},

{D C}^{2} = {B D}^{2} = 2 A B \times A D

∠

90^{\circ},

D C^{2} - B D^{2} =

△ A B C

∠ A = 90^{0}, C A = A B

D

A B

D C^{2} - B D^{2} = 2 A B . A D

In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

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