Question

In given fig. 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 X'Y' at B. Prove that angle aob=90 degree

Answer 1

join OC
in APO and ACO,
AP=AC [Tangents from same exterior point]
AO=AO [Common]
OP=PC [Radius]
this, by SSS criteria, both the triangles are congruent
angle AOP=angle AOC [CPCT]
Similarly BCO IS CONGRUENT TO BQO
And angle BOC = angle BOQ by CPCT.
PQ is a straight line passing through center such that
anglePOA+ angleAOC+ angleCOB+ angleQOB=180degree
2 angle POC+ 2angle COB=180 [ angle AOP = angle AOC and angle BOC = angle BOQ as proved earlier ]
2( angle POC+ angle COB)=180
angle POC+ angle COB=180 /2
angle POC+ angle COB=90 degree
angle AOB= 90 degree.
HENCE PROVED.



hope it helped u......

Answer 2

Join OC

In ΔOPA and ΔOCA

OP = OC (radii of same circle)

PA = CA (length of two tangents from an external point)

AO = AO (Common)

ΔOPA ≅ ΔOCA  (By SSS congruency criterion)

Hence, ∠1 = ∠2 (cpct)

Similarly ∠3 = ∠4

Now,

∠PAB + ∠QBA = 180°

⇒ 2∠2 + 2∠4 = 180°

⇒ ∠2 + ∠4 = 90°

⇒ ∠AOB = 90° (Angle sum property)