Let PQR be an acute-angled triangle in which | vedclass

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Question

(A) In $	riangle PQR$,let $PQ = r$,$QR = p$,and $PR = q$. Since $PQ < QR$,we have $r < p$.$1$. The altitude $QQ_1$ is the shortest segment from $Q$ t

Vedclass · Accepted Answer

$PQ_1 < PQ_2 < PQ_3$

Vedclass · Answer

(A) In $	riangle PQR$,let $PQ = r$,$QR = p$,and $PR = q$. Since $PQ $1$. The altitude $QQ_1$ is the shortest segment from $Q$ to the line $PR$. Thus,$PQ_1$ is the smallest distance.$2$. The median $QQ_3$ bisects $PR$,so $PQ_3 = \frac{1}{2} PR$.$3$. By the Angle Bisector Theorem,the angle bisector $QQ_2$ divides $PR$ in the ratio of the adjacent sides $PQ:QR = r:p$. Thus,$PQ_2 = \left(\frac{r}{r+p}ight) PR$.Since $r  2r$,which implies $\frac{r}{r+p} Therefore,$PQ_2 Comparing the positions of the feet on $PR$,we have $PQ_1 < PQ_2 < PQ_3$. The correct option is $A$.

Let $PQR$ be an acute-angled triangle in which $PQ < QR$. From the vertex $Q$,draw the altitude $QQ_1$,the angle bisector $QQ_2$,and the median $QQ_3$,with $Q_1, Q_2, Q_3$ lying on $PR$. Then,

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