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 ïƒAOB = 90Â°
In Î”APO and Î”ACO
AP = AC (tangents from a point)
OP = OC (radii)
OA = OA (common side)
Hence; Î”APO â‰ˆ Î”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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