In figure $D E F G$ is a square and $∠ B A C = 90^{o}$ . Show that ${D E}^{2} = B D \times E C$

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Solution

Given: $D E F G$ is a square and $∠ B A C = 90 °$ .

To Prove: $D E ² = B D \times E C$ .

Proof :
In $Δ A F G a n d Δ D B G$
$∠ G A F = ∠ B D G [90 °]$
$∠ A G F = ∠ D B G$ [corresponding angles because GF|| BC and AB is the transversal]
$Δ A F G \approx Δ D B G$ [by AA Similarity Criterion] …………(1)

In $Δ A G F & Δ E F C$
$∠ A F G = ∠ C E F$ $[90 °]$
$∠ A F G = ∠ E C F$ [corresponding angles because GF|| BC and AC is the transversal]
$Δ A G F \approx Δ E F C$ [by AA Similarity Criterion] …………(2)

From equation 1 and 2.

$Δ D B G \approx Δ E F C$ $B D / E F = D G / E C$
$B D / D E = D E / E C$ [ DEFG is a square]
$D E^{2} = B D \times E C$ .
Proved.

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In the adjoining figure, □ DEFG is a square and ∠ BAC = 90 ° . Prove that (i) △ AGF ~ △ DBG (ii) △ AGF ~ △ EFC (iii) △ DBG ~ △ EFC (iv) {DE}^{2} = BD . EC

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In figure DEFG is a square and ∠BAC=90o. Show that DE2=BD×EC

In figure $D E F G$ is a square and $∠ B A C = 90^{o}$ . Show that ${D E}^{2} = B D \times E C$