Q. 215.0( 1 Vote )

In Fig. 2 DE || A

Answer :

DE || AC, By basic proportionality theorem


[1]


Also, DC || AP


[2]


From [1] and [2], we get



Hence, proved!


OR


Let TP and TQ are two tangents of a circle at points P and Q respectively with center O.



To prove: PTQ = 2OPQ


Let PTQ = θ


As lengths of tangents drawn from an external point to the circle are equal, therefore TP = TQ.


ΔPQT is an isosceles triangle.


TPQ = TQP = 1/2(180° - θ) = 90° – 1/2θ


Also, tangent at any point of a circle is perpendicular to the radius through the point of contact.


OPQ = OPT – TPQ = 90° – (90°– 1/2θ)


= 1/2θ


= 1/2PTQ


Thus, PTQ = 2OPQ.


Hence, proved.


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