# If S is a point on side PQ of a Δ PQR

Question:

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

(a) $P R \cdot Q R=R S^{2}$

(b) $Q S^{2}+R S^{2}=Q R^{2}$

(c) $P R^{2}+Q R^{2}=P Q^{2}$

(d) $P S^{2}+R S^{2}=P R^{2}$

Solution:

(c) Given, in $\triangle P Q R$,

$P S=Q S=R S$ $\ldots$ (i)

$\ln \Delta P S R_{1}$ $P S=R S$ [from Eq. (i)]

$\Rightarrow$  $\angle 1=\angle 2$ $\ldots$ (ii)

Similarly, in $\Delta R S Q$,

$\Rightarrow \quad \angle 3=\angle 4$ .....(iii)

[corresponding angles of equal sides are equal]

Now, in $\triangle P Q R$, sum of angles $=180^{\circ}$

$\Rightarrow \quad \angle P+\angle Q+\angle R=180^{\circ}$  [using Eqs. (ii) and (iii)]

$\Rightarrow \quad \angle 2+\angle 4+\angle 1+\angle 3=180^{\circ}$

$\Rightarrow \quad \angle 1+\angle 3+\angle 1+\angle 3=180^{\circ}$

$\Rightarrow \quad 2(\angle 1+\angle 3)=180^{\circ}$

$\Rightarrow \quad \angle 1+\angle 3=\frac{180^{\circ}}{2}=90^{\circ}$

$\therefore \quad \angle R=90^{\circ}$

In $\Delta P Q R$, by Pythagoras theorem,

$P R^{2}+Q R^{2}=P Q^{2}$