Let $P Q R$ be an acute-angled triangle in which $P Q < Q R$ . From the vertex $Q$ draw the altitude $Q Q_{1}$ , the angle bisector $Q Q_{2}$ and the median $Q Q_{3}$ , with $Q_{1}, Q_{2}, Q_{3}$ lying on $P R$ . Then

P Q_{1} < P Q_{2} < P Q_{3}

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P Q_{2} < P Q_{1} < P Q_{3}

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P Q_{1} < P Q_{3} < P Q_{2}

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P Q_{3} < P Q_{1} < P Q_{2}

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Solution

The correct option is A $P Q_{1} < P Q_{2} < P Q_{3}$
Given :
$P Q < Q R$

As we know $Q Q_{3}$ is a median, so
$P Q_{3} = Q_{3} R \Rightarrow P Q_{3} = \frac{1}{2} P R \dots (i)$
Using property of angle bisector,
$P Q_{2} : Q_{2} R = P Q : Q R \Rightarrow P Q_{2} = \frac{P Q}{(P Q + Q R)} P R$

$∴ P Q_{2} < \frac{1}{2} P R$ .....(ii) [ $∵ P Q < Q R$ ]
In $△ P Q Q_{2} and △ R Q Q_{2}$
$∠ P Q Q_{2} + ∠ P Q_{2} Q + ∠ Q P Q_{2} = ∠ R Q Q_{2} + ∠ R Q_{2} Q + ∠ Q R Q_{2}$
$\Rightarrow ∠ P Q_{2} Q + ∠ Q P Q_{2} = ∠ R Q_{2} Q + ∠ Q R Q_{2}$
[ $∵ Q Q_{2}$ is angle bisector ]
As we know $P Q < Q R$ so $∠ Q P Q_{2} > ∠ Q R Q_{2}$
$∴ ∠ P Q_{2} Q < ∠ R Q_{2} Q$
$∠ R Q_{2} Q > 90^{\circ}$ [ $∵ ∠ P Q_{2} Q + ∠ R Q_{2} Q = 180^{\circ}$ ]
so $P Q_{1}$ lies inside the $△ P Q Q_{2}$

Hence, $P Q_{1} < P Q_{2} < P Q_{3}$

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Let PQR be an acute-angled triangle in which PQ<QR. From the vertex Q draw the altitude QQ1, the angle bisector QQ2 and the median QQ3, with Q1,Q2,Q3 lying on PR. Then

Let $P Q R$ be an acute-angled triangle in which $P Q < Q R$ . From the vertex $Q$ draw the altitude $Q Q_{1}$ , the angle bisector $Q Q_{2}$ and the median $Q Q_{3}$ , with $Q_{1}, Q_{2}, Q_{3}$ lying on $P R$ . Then