If S is a point on side PQ of a triangle PQR | vedclass

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(B) Given,in $	riangle PQR$,$PS = QS = RS$ ... $(1)$In $	riangle PSR$,$PS = RS$ (from Eq. $1$),therefore $\angle 1 = \angle 2$ ... $(2)$ (angles opp

Vedclass · Accepted Answer

$PR^2 + QR^2 = PQ^2$

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(B) Given,in $	riangle PQR$,$PS = QS = RS$ ... $(1)$In $	riangle PSR$,$PS = RS$ (from Eq. $1$),therefore $\angle 1 = \angle 2$ ... $(2)$ (angles opposite to equal sides are equal).Similarly,in $	riangle RSQ$,$RS = QS$ (from Eq. $1$),therefore $\angle 3 = \angle 4$ ... $(3)$ (angles opposite to equal sides are equal).In $	riangle PQR$,the sum of angles is $180^{\circ}$.$\angle P + \angle Q + \angle R = 180^{\circ}$$\angle 2 + \angle 4 + (\angle 1 + \angle 3) = 180^{\circ}$Substituting $\angle 2 = \angle 1$ and $\angle 4 = \angle 3$ from $(2)$ and $(3)$:$\angle 1 + \angle 3 + \angle 1 + \angle 3 = 180^{\circ}$$2(\angle 1 + \angle 3) = 180^{\circ}$$\angle 1 + \angle 3 = 90^{\circ}$Thus,$\angle R = 90^{\circ}$.In $	riangle PQR$,by Pythagoras theorem,$PR^2 + QR^2 = PQ^2$.

If $S$ is a point on side $PQ$ of a $\triangle PQR$ such that $PS = QS = RS$,then

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