Question

In the given figure, P and Q be the centres of two intersecting circles and AB//PQ.

Prove that AB=2PQ.

Guys Ive done construction here by mysel. Hope its correct!!

Answer 1

First.. Join points P to X, P to Y.. The Q to X, Q to y this makes 4 triangle (POX, POY, QOX, QOY).. where the intersecting point of pq and xy is o.. Now Prove triangle POX Is congruent to Triangle POY.. By CPCT:OX=OY...now join K to O.. This makes 2 triangles prove the congruence between triangle PKX And triangle POX so by CPCT OX =KX.. So 2 KX =AX.. lly 2XL =XB... As PO=KX AND QO =XL.. we can take 2 PQ =AB

Answer 2

SOLUTION____(◠‿◕)

To prove: AB=2PQ

Given: Two circle with P and Q and AB||PQ

Proof: Draw PK perpendicular AX and also AK=KY

similarly, QL perpendicular to BX also BL=LX

Now, PQ and KL is a rectangle.

so, PQ||KL and PK||QL {opposite side parallel}

and, PQ' KL is a ||gm.

But a ||gm with are right angle is a rectangle. Thus, PQ and KL is a rectangle.

So, PQ =KX+KL {opposite sides} and multiplying by 2 on both sides.

=2PQ= 2KX+2KL

= 2PQ=AX+XB

= 2PQ= AB

HENCE proved..(◍•ᴗ•◍)❤