In figure below, if $AB\ ⊥\ BC$, $DC\ ⊥\ BC$, and $DE\ ⊥\ AC$, prove that $Δ\ CED\ ∼\ Δ\ ABC$.

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**Given:**

In the given figure $AB\ âŠ¥\ BC$, $DC\ âŠ¥\ BC$, and $DE\ âŠ¥\ AC$.

**To do:**

We have to prove that $Î”\ CED\ ∠¼\ Î”\ ABC$.

**Solution:**

In $\vartriangle ABC$ and $\vartriangle CED$,

$\angle BAC+\angle BCA=90^o$

$\angle BCA+\angle ECD=90^o$

This implies,

$\angle BAC=\angle ECD$

In $\vartriangle ABC$ and $\vartriangle CED$,

$\angle ABC=\angle CED=90^o$

$\angle BAC=\angle ECD$

Therefore,

$Î”\ CED\ ∠¼\ Î”\ ABC$.

Hence proved.