In the given Fig., DEFG is a square and ∠BAC=90°. Prove that (i) ΔAGF~ΔDBG (ii) ΔAGF~ΔEFG (iii) ΔDBG

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In the given Fig., DEFG is a square and ∠BAC=90°. Prove that
(i) ΔAGF~ΔDBG (ii) ΔAGF~ΔEFG (iii) ΔDBG~ΔEFC (iv) DE^2=BD×EC
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By - Vineet Pandey
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Sir Lekin question mai toh bola hei nahi gya hai ki GF is parallel to BC ?

Aakirtidwivedi
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Wawawaww kya batt hay.boqrd kya hia???🥴🥴

prabinbhatt