Question

O is a point on side PQ of a APQR such that PO = QO = RO, then (a) RS² = PR × QR (b) PR² + QR² = PQ² (c) QR² = QO² + RO² (d) PO² + RO² = PR²

Answer 1

In ΔPQR , 

PO=OQ=RO (given) 

Now , in ΔPSR , 

PO=OR (given) 

∴∠1=∠P [Angles opposite to equal sides in a triangle are equal]

Similarly , in ∠ORQ , 

RO=OQ (given) 

∠Q=∠2 

Now , in ΔPQR , 

∠P+∠Q+∠PRQ=180^∘

  [By Angle sum property of a triangle] 

⇒∠1+∠2+(∠1+∠2)=180^∘

  ⇒2(∠1+∠2)=180^∘

  ⇒∠1+∠2=90^∘

  ⇒∠PRQ=90^∘

By Pythagoras theorem , we have 

PQ²=QR²+PQ²

option B is correct