Answers

2016-02-12T19:36:55+05:30
In triangles OPC and OBC,
=> CO = CO (common)
=> angle OPC = angle OBC = 90 (both are angles formed by radius-tangent intersection)
=> OP = OB (radii of same circle)
Therefore, triangle OPC is congruent to triangle OBC, by RightAngle-Hypotenuse-Side test.
=> angle PCO = angle BCO (CPCT) ...(1)

Similarly prove triangles AQO and ABO congurent.
=> angle OAQ = angle OAB (CPCT) ...(2)

In quadrilateral PCAQ,
angle OPC = angle OQA = 90
angles PCA+CAQ+AQP+QPC = 360 (sum of interior angles)
=> (PCO+BCO)+(OAB+OAQ)+90+90 = 360
=> 2BCO+2OAB+180 = 360 (From 1 and 2)
=> 2(BCO+OAB) = 360-180
=> BCO+OAB = 180/2 = 90 ...(3)

Now, in triangle AOC,
angles AOC+BCO+OAB = 180
=> AOC+90 = 180 (From 3)
=> AOC = 180-90 = 90

Thus Proved.
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I didn't use the same point names, but think of it as a general proof
ok thNX
No problem!