Question

IF S is a point on side PQ of a triangle PQR such that PS=QS=RS , then Prove that PR2 + QR2 = PQ2

Answer 1

The answer is in the attachment.... Refer to it
...............

Answer 2

Hence proved.

Step-by-step explanation:

Given,

In $\triangle PQR$, $S$ is a point on $PQ$ such that $PS=QS=RS$ the prove $PR^{2} +QR^{2} =PQ^{2}$

∴$\angle1=\angle2=\angle3=\angle4$ (∵$PS=QS=RS$ )

From $\triangle PQR$ ,

$\angle1+\angle2+\angle3+\angle4=180$°

⇒$4\angle1=180$°

⇒$\angle 1=45$°

∴$\angle R=\angle1+\angle2=45+45=90$°

∴ $\triangle PQR$ is a right angle triangle where $\angle R=90$°

So,

From Pythagorean theorem,

$PR^{2} +QR^{2} =PQ^{2}$ where $PQ=Hypotenuse$

Hence proved.