In the given figure $P Q = P R$ , then prove that $P S > P Q$

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Solution

$∠ 1 = ∠ 2 (g i v e n)$
$P R > P Q (g i v e n)$
$I n Δ P Q R$
$P R > P Q (g i v e n)$
$\Rightarrow ∠ P Q R > ∠ P R Q + ∠ R P S (sin c e ∠ Q P S = ∠ R P S)$
$\Rightarrow - (∠ P Q R + ∠ Q P S) < - (∠ P R Q + ∠ R P S)$
$\Rightarrow 180 - (∠ P Q R + ∠ Q P S) < 180 - (∠ P R Q + ∠ R P S)$
$\Rightarrow 180 - (∠ P Q S + ∠ Q P S) < 180 - (∠ P R S + ∠ R P S)$
$\Rightarrow ∠ P S Q < ∠ P S R$ (by angle sum property)
$\Rightarrow ∠ P S R > ∠ P S Q$

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In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.

Question 5
In the given figure, PR > PQ and PS bisects $∠$ QPR. Prove that $∠$ PSR > $∠$ PSQ
.

P R > P Q

P S

∠ Q P R

∠ P S R > ∠ P S Q

∠ P Q R = ∠ P R S

P Q R

P R S

P R = 8 c m, P S = 4 c m

P Q

≅

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