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

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Solution

$(i) S eg GF ∥ S eg DE [O pposite sides of the square] S eg GF ∥ S eg BC In △ AGF & △ DBG, ∠ GAF = ∠ BDG [90 ° each] ∠ AGF = ∠ DBG [Corresponding angle s, as GF ∥ BC] By AA test of similarity, △ AGF ~ △ DBG (ii) In △ AGF & △ EFC, ∠ GAF = ∠ FEC [90 ° each] ∠ AFG = ∠ ECF [Corresponding angle s, as GF ∥ BC] By AA test of similarity, △ AGF ~ △ EFC (iii) △ AGF ~ △ EFC and △ AGF ~ △ DBG . Therefore, △ DBG ~ △ EFC . (iv) \frac{BD}{FE} = \frac{DG}{EC} [Corresponding sides of similar triangles] \Rightarrow DG \times FE = BD \times EC B ut DG = EF = DE [Sides of the square] \Rightarrow DE \times DE = DB \times EC \Rightarrow {DE}^{2} = DB . EC$

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Similar questions

In figure, DEFG is a square and $∠ B A C = 90^{o}$ , Prove that
(i) $△ A G F \sim △ D B G$
(ii) $△ A G F \sim △ E F C$
(iii) $△ D B G \sim △ E F C$
(iv) ${D E}^{2} = B D \times E C$

Q. In given figure, DEFG is a square and

∠ B A C = 90^{0}

. Prove that

(i)

△ A G F \sim △ D B G

(ii)

△ A G F \sim △ E F C

(iii)

△ D B G \sim △ E F C

(iv)

{D E}^{2} = B D \times E C

$In the given figure, DEFG is a square and ∠ B A C = 90^{o} .$

$Which of the following is true?$

Q. In figure,

D E F G

is a square and

∠ B A C = 90

. Show that

D E^{2} = B D \times E C

Q. 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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In the adjoining figure, □DEFG is a square and ∠BAC=90°. Prove that(i) △AGF~△DBG(ii) △AGF~△EFC(iii) △DBG~△EFC(iv) DE2=BD.EC

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