ABD - PQS and AD and PS are the medians- then AC BC - QR PR-AC BC - PR QR-AB AC - QR PR-AC AB - QR PR-

The correct option is B AC BC = PR QR If ADB ∼ PSQ , AB PQ = AD PS =BD QS (corresponding sides of similar triangles will be in same ratio) ∠ B = ∠ Q (corresp ...

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Byju's Answer

Standard VIII

Mathematics

SAS similarity

ABD ∼ PQS and...

Question

$△ A B D \sim △ P Q S$ and AD and PS are the medians, then

$\frac{A C}{B C} = \frac{P R}{Q R}$

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$\frac{A C}{B C} = \frac{Q R}{P R}$

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$\frac{A C}{A B} = \frac{Q R}{P R}$

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$\frac{A B}{A C} = \frac{Q R}{P R}$

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Solution

The correct option is A
$\frac{A C}{B C} = \frac{P R}{Q R}$

If $△ A D B \sim △ P S Q$ ,

$\frac{A B}{P Q} = \frac{A D}{P S} = \frac{B D}{Q S}$
(corresponding sides of similar triangles will be in same ratio)

$∠ B = ∠ Q$ (corresponding angles of similar triangles will be equal)

$\Rightarrow \frac{A B}{P Q} = \frac{2 B D}{2 Q S} = \frac{B C}{Q R}$

Now, consider $△ A B C$ and $△ P Q R$ .

$∠ B = ∠ Q$
$\Rightarrow \frac{A B}{P Q} = \frac{B C}{Q R}$

$∴ △ A B C \sim △ P Q R$ (by SAS similarity criterion)

$\frac{A C}{P R} = \frac{B C}{Q R} = \frac{A B}{P Q}$
$\Rightarrow \frac{A C}{B C} = \frac{P R}{Q R}$

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Similar questions

In the following figure, the sides AB and BC and the median AD of triangle ABC are respectively equal to the sides PQ and QR and median PS of the triangle PQR. Prove that $Δ$ PQR are congruent.