Question

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5. Prove that the perpendicular at the point of contact to the tangent to a circle pare
through the centre.​

Answer 1

Answer:

hope that it will help u ....

Step-by-step explanation:

Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

OA PA and OB PB ... (1)

In OPA and OPB:

OAP = OBP (Using (1))

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore, OPA OPB (RHS congruency criterion)

PA = PB

(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.