Question

In ∆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:
(i) ∆OMA ∼ ∆OLC
(ii) OA/OC = OM/OL

Answer 1

SOLUTION :  

GIVEN : In ΔABC, AL ⊥BC and CM ⊥AB.AL and CM intersect at O.

TO PROVE :

(i)  ΔOMA∼ΔOLC

(ii) OAOC=OMOL

PROOF :  

(i) In ΔOMA and ΔOLC,

∠AOM = ∠COL  (vertically opposite angles)

∠OMA = ∠OLC  (each 90°)

ΔOMA∼ΔOLC   (By A-A similarity)

(ii) Since, ΔOMA∼ΔOLC

Then, OM/OL = OA/OC = MA/LC  

(The corresponding sides of similar triangles are proportional)

OA/OC = OM/OL

Hence proved.

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Answer 2

$corresponding \: sides \: of \: two \: \\ similar \: triangle \: are \: proportional$