In figure below, $\triangle ABC$ is right angled at C and $DE \perp AB$. Prove that $\triangle ABC \sim\ \triangle ADE$ and hence find the lengths of $ A E $ and $ D E $.
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Given:

In the given figure $\triangle ABC$ is right angled at C and $DE \perp AB$.
To do:

We have to prove that $\triangle ABC \sim\ \triangle ADE$ and hence find the lengths of \( A E \) and \( D E \).

Solution:

$\triangle ABC$ is right angled at C. Therefore,

By Pythagoras theorem,

$AB^2=AC^2+BC^2$

$AB^2=(12)^2+(5)^2$

$AB^2=144+25$

$AB=\sqrt{169}$

$AB=13\ cm$

In $\triangle ACB$ and $\triangle AED$

$\angle ACB = \angle AED = 90^o$

$\angle BAC = \angle EAD$   (Common angle)

Therefore,

$\triangle ACB \sim\ \triangle AED$    (By AA similarity)

This implies,

$\frac{AC}{AE} = \frac{AB}{AD}=\frac{CB}{ED}$    (Corresponding sides are proportional)

$\frac{5}{AE} = \frac{13}{3}=\frac{12}{ED}$

$AE=\frac{5\times3}{13}=\frac{15}{13}\ cm$

$DE=\frac{12\times3}{13}=\frac{36}{13}\ cm$

The lengths of \( A E \) and \( D E \) are $\frac{15}{13}\ cm$ and $\frac{36}{13}\ cm$ respectively.

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Updated on: 10-Oct-2022

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