# Circle

## Exercise 10.2 Part 4

Question: 8 - A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC

Answer: Construction: Draw a circle with centre O. Draw a quadrilateral ABCD which touches the circle at P, Q, R and S.

To Prove: AB + CD = AD + BC AP = AS

BP = BQ

CQ = CR

DR = DS

(Tangents from same external point are equal)

Adding all the four equations from above; we get:

AP + BP + CR + DR = AS + DS + BQ + CQ

Or, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

Or, AB + CD = AD + BC proved

Question: 9 - In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90° In ΔAPO and ΔACO

AP = AC (tangents from a point)

OP = OC (radii)

OA = OA (common side)

Hence; ΔAPO ≅ ΔACO

So, ∠PAO = ∠CAO

Hence; AO is the bisector of ∠PAC.

Similarly, it can be proved that

BO is the bisector of ∠QBC

Now, XY || X’Y’ (given)

So, ∠AOB = Right angle

(Bisectors of internal angles on one side of transversal intersect at right angle.)