In $\triangle ABC$, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that: $\triangle OMA \sim OLC$.

Given:

In $\triangle ABC$, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. AL and CM intersect at O.
To do:

We have to prove that: $\triangle OMA \sim OLC$.
Solution:

$AL \perp BC$ and $CM \perp AB$

In $\triangle OMA$ and $\triangle OLC$

$\angle OMA=\angle OLC=90^o$

$\angle MOA=\angle LOC$  (Vertically opposite angles)

Therefore,

$\triangle OMA \sim\ \triangle OLC$   (By AA similarity)

Hence proved.

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Updated on: 10-Oct-2022

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