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# there is a cicle with centre o.a triangle ABP is inscribed in it such that the tangent TP at the point P is parallel to AB.prove that APB is an isosceles triangle.

Asked by 1st March 2013, 4:26 PM
Join OP and extend it to intersect AB at point Q. Since, PT is the tangent, angle(OPT) = 90. Also, since PT is parallel to AB, so angle(OQB) = angle (OQA) = 90

Now, in triangle(AOQ) and triangle (BOQ)
angle(OQB) = angle (OQA) (90 degree each)
OQ = QO (common)
triangle(AOQ) is congruent to triangle (BOQ) (by RHS)
Hence, AQ = QB  (By CPCT)

Now in triangle(APQ) and triangle (BPQ)
angle(PQB) = angle (PQA) (90 degree each)
PQ = QP (common)
AQ = QB  (from above)
Hence triangle(APQ) is congruent to triangle (BPQ) (by SAS)

So, angle (PAQ) = angle (PBQ) (By CPCT)
Since, sides opposite to equal angles are equal in a triangle
So, AP = PB
Hence, triangle APB is an isosceles triangle.
Answered by Expert 3rd March 2013, 6:48 AM
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