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

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Solution

(i) In $△ A G F a n d △ D B G$ , we have
$∠ G A F = ∠ B D G$ [Each equal to $90^{o}]$
and, $∠ A G F = ∠ D B G$ [Corresponding angles]
$∴ △ A G F \sim △ D B G$ [By AA-criterion of similarity]

(ii) In $△ A G F a n d △ E F C$ , we have
$∠ F A G = ∠ C E F$ [Each equal to $90^{o}]$
and, $∠ A F G = ∠ E C F$
$∴ △ D B G \sim △ E F C$ [By AA-criterion of similarity]
(iii)Since $△ A G F \sim △ D B G a n d △ A G F \sim △ E F C$
$∴ △ D B G \sim △ E F C$
(iv)We have, $△ D B G \sim △ E F C [U s i n g (i i i)]$
$∴ \frac{B D}{E F} = \frac{D G}{B C}$
$△ \frac{B D}{D E} = \frac{D E}{E C}$
$[∵ D E F G i s a s q u a r e ∴ E F = D E, D G = D E]$
$\Rightarrow {D E}^{2} = B D \times B C$

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∠ B A C = 90^{0}

△ A G F \sim △ D B G

△ A G F \sim △ E F C

△ D B G \sim △ E F C

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

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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$Which of the following is true?$

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