$△ P Q R$ is an isosceles triangle with $P Q = P R$ , $Q P$ is produced to $S$ and $P T$ bisects the extension angle $2 x^{o}$ . Prove that $∠ Q = x^{o}$ and hence prove that $P T ∥ Q R$ .

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Solution

Given: $△ P Q R$ is an isosceles triangle with $P Q = P R$
Proof: $P T$ bisects exterior angle $∠ S P R$ and therefore $∠ S P T = ∠ T P R = x^{o}$
$∴$ $∠ Q = ∠ R$ (Property of an isosceles triangle)
also we know that in any triangle,
exterior angle $=$ sum of the interior opposite angles.
$∴$ In $△ P Q R$ , Exterior angle $∠ S P R = ∠ P Q R + ∠ P R Q$
$2 x^{o} = ∠ Q + ∠ R$
$= ∠ Q + ∠ Q$
$2 x^{o} = 2 ∠ Q$
$x^{o} = ∠ Q$
To prove: $P T ∥ Q R$
Lines $P T$ and $Q R$ are cut by the transversal $S Q$ . We have $∠ S P T = x^{o}$
Hence, $∠ S P T$ and $∠ P Q R$ are corresponding angles: $P T ∥ Q R$

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Q. If

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\frac{P K}{K S} = \frac{P T}{T R}

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PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.

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△ P Q R

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8 {P T}^{2} = 3 {P R}^{2} + 5 {P S}^{2}

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△PQR is an isosceles triangle with PQ=PR, QP is produced to S and PT bisects the extension angle 2xo. Prove that ∠Q=xo and hence prove that PT∥QR.

$△ P Q R$ is an isosceles triangle with $P Q = P R$ , $Q P$ is produced to $S$ and $P T$ bisects the extension angle $2 x^{o}$ . Prove that $∠ Q = x^{o}$ and hence prove that $P T ∥ Q R$ .