ABC is right-angled at A and DEFG is a square as shown in the figure. Then:

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Solution

Q.
$Δ A B C$ is right-angled at A and DEFG is a square as shown in the figure. Then:

$(i) Δ C B G ~ Δ E F C (i i) B D \times E C = D E ²$

$(i i i) Δ A G F ~ Δ D B G (i v) Δ A G F ~ Δ E F C$

Choose the correct option from the following:

1 (iii) and (iv) are correct.

2 (i), (ii) and (iii) are correct.

3 (ii), (iii) and (iv) are correct.

4 (i) and (ii) are correct.

Ans.
(3)

$I n Δ A G F a n d Δ D B G :$

$∠ G A F = ∠ B D G = 90 ° a n d$

$∠ A G F = ∠ D B G (C o r r e s p o n d i n g a n g l e s)$

$∴ Δ A G F ~ Δ D B G (B y A A s i m i l a r i t y) \dots (A) ∴ (i i i) i s c o r r e c t .$

$S i m i l a r l y, i n Δ A G F a n d Δ E F C :$

$∠ G A F = ∠ C E F = 90 ° a n d$

$∠ A F G = ∠ F C E (C o r r e s p o n d i n g a n g l e s)$

$∴ Δ A G F ~ Δ E F C (B y A A s i m i l a r i t y) \dots (B) ∴ (i v) i s c o r r e c t .$

$F r o m (A) a n d (B) : Δ D B G ~ Δ E F C$

$\frac{D B}{E F} = \frac{D G}{E C}$

$B u t E F = D G = D E (s i d e o f s q u a r e)$

$\frac{D B}{D E} = \frac{D E}{E C} \Rightarrow D E^{2} = D B \times E C ∴ (i i) i s c o r r e c t .$

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