Given: DEFG is a square and ∠BAC = 90°. To Prove: DE² = BD × EC. Proof :
In ∆ AFG & ∆DBG ∠GAF = ∠BDG [ 90°] ∠AGF = ∠DBG [corresponding angles because GF|| BC and AB is the transversal] ∆AFG ~ ∆DBG [by AA Similarity Criterion] …………(1)
In ∆ AGF & ∆EFC ∠AFG = ∠CEF [ 90°] ∠AFG = ∠ECF [corresponding angles because GF|| BC and AC is the transversal] ∆AGF ~ ∆EFC [by AA Similarity Criterion] …………(2)
From equation 1 and 2.
∆DBG ~ ∆EFC BD/EF = DG /EC BD/DE = DE /EC [ DEFG is a square]
DE² = BD × EC .
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FIGURE IS IN THE ATTACHMENT.Given: DEFG is a square and ∠BAC = 90°.
To Prove: DE² = BD × EC.
Proof :
In ∆ AFG & ∆DBG
∠GAF = ∠BDG [ 90°]
∠AGF = ∠DBG [corresponding angles because GF|| BC and AB is the transversal]
∆AFG ~ ∆DBG [by AA Similarity Criterion] …………(1)
In ∆ AGF & ∆EFC
∠AFG = ∠CEF [ 90°]
∠AFG = ∠ECF [corresponding angles because GF|| BC and AC is the transversal]
∆AGF ~ ∆EFC [by AA Similarity Criterion] …………(2)
From equation 1 and 2.
∆DBG ~ ∆EFC
BD/EF = DG /EC
BD/DE = DE /EC [ DEFG is a square]
DE² = BD × EC .
HOPE THIS WILL HELP YOU....
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