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In the given ...

Question

In the given figure, $A B D$ is a triangle right angled at $A$ and $A C$ perpendicular to $B D$ .Show that ${A C}^{2} = B C . D C$

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Solution

In right angled triangle $A C B$ and $D C A$
$∠ A C B = ∠ D C A = 90^{\circ}$
$∠ B A C = ∠ A D C$
$∴ △ A C B \sim △ D C A$ using AA similar condition
$∴ \frac{A C}{D C} = \frac{B C}{A C}$
$∵$ corresponding sides of two similar triangles are proportional
$\Rightarrow {A C}^{2} = B C \times D C$

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Similar questions

A B D

A

A C

⊥

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

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

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

A C ⊥ B D

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

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

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

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

△ A B D

A C ⊥ B D .

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

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

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

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

In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC × BD

(ii) AC² = BC × DC

(iii) AD² = BD × CD

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