Q. 215.0( 1 Vote )

# In Fig. 2 DE || A

Answer :

DE || AC, By basic proportionality theorem Also, DC || AP From  and , 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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