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 $. "
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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