The vertices of square DEFG are on the sides of triangle ABC angle A=90 then prove that DE square=BD×EC
surajsingh01082004:
can u get me the diagram....
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Given: ABC is a triangle in which ∠BAC = 90° and DEFG is a square.
To prove: DE2 = BD x EC.
Proof: In △AGF and △DBG,
∠AGF = ∠GBD (corresponding angles)
∠GAF = ∠BDG (each = 90‘)
∴△AGF ~ △DBG .....(i)
Similarly, △AFG ~ △ECF (AA Similarity).....(ii)
From (i) and (ii), △DBG ~ △ECF.
BD/EF - BG/FC - DG/EC
BD/EF - DG/EC
EF × DG = BD × EC......(iii)
Also DEFG is a square
⇒ DE = EF = FG = DG .....(iv)
From (iii) and (iv),
DE2 = BD × EC. _____________proved
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