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: $\frac{OA}{OC}=\frac{OM}{OL}$.
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 $\frac{OA}{OC}=\frac{OM}{OL}$.
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)
This implies,
$\frac{OA}{OC}=\frac{OM}{OL}$ (Corresponding parts of similar triangles are proportional)
Hence proved.
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