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Construction of a Similar Triangle, When One Vertex Is Common

Question 4In ...

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Question 4
In the figure,
$\frac{Q R}{Q S} = \frac{Q T}{P R} a n d$
$∠ 1 = ∠ 2$ .
$S h o w t h a t Δ P Q S \sim Δ T Q R$ .

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Solution

$I n Δ P Q R, ∠ P Q R = ∠ P R Q (G i v e n)$

$∴ P Q = P R . . . (i)$ (sides opposite to equal angles are equal)

$G i v e n, \frac{Q R}{Q S} = \frac{Q T}{P R}$

Using ( $i$ ), we get
$\frac{Q R}{Q S} = \frac{Q T}{Q P} . . . (i i)$

In $Δ P Q S$ and $Δ T Q R$ ,
$\frac{Q R}{Q S} = \frac{Q T}{Q P}$ [using ( $i i$ )]

$∠ Q = ∠ Q$ (Common)

$∴ Δ P Q S \sim Δ T Q R$ [SAS similarity criterion]

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Construction of a Similar Triangle, When One Vertex Is Common

Standard X Mathematics

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