In ΔABC, 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 :
(i) ΔOMA~ ΔOLC
(ii) OA/OC = OM/OL
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Hence it is proved that,
(i) ΔOMA~ ΔOLC
(ii) OA/OC = OM/OL
Consider the figure while going through the following steps:
Given,
In ΔABC, AL and CM are the ⊥s from the vertices A and C to BC and AB.
AL and CM intersect at O
⇒ AL ⊥BC
CM ⊥ AB
In Δ OMA and Δ OLC
∠ AMO = ∠ CLO (angles are equal to 90°)
∠ MOA = ∠ LOC (vertically opposite angles)
By AA similarity, we have,
∴ Δ OMA ~ Δ OLC
As corresponding sides of similar triangles are proportional, we have,
∴ OA/OC = OM/OL
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