Tangent to Circle
NCERT 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°
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.)