# Circles

## Soution of NCERT Exercise 10.2 - part - 4

Exercise 10.2

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 �ƒAOB = 90° In �”APO and �”ACO

AP = AC (tangents from a point)

OA = OA (common side)

Hence; �”APO �‰ˆ �”ACO

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.

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