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°

**Answer:**

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.)

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