Question

0.22. Prow that the lengths of the tangents drawn from the external point to a circle are
सिद्ध कीजिए कि किसी बाह्य बिंदु से वृत्त पर खींची गई स्पर्श रेखायें समान लम्बाई की होती
है।​

Answer 1

Step-by-step explanation:

ਰਾਠ-1 ਸਾਧਨ-ਪ੍ਕਸਮਿ ਅਤੇ ਸਿਕ- ਸੰ ਕਾਲ

Answer 2

Given:

A circle with center O and PA and PB are two tangents to the circle drawn from an external point P.

To Find:

We have to prove that the length of the tangents PA and PB are equal.

Solution:

Draw a circle with two tangents touching it with the same external point as shown below. Join OA, OB, and OP.

Tangent is perpendicular to radius: A tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.

OA ⊥ PA

OB ⊥ PB

RHS criterion for Congruence: It states that a triangle is said to be a congruent triangle if the length of the hypotenuse and one side of one triangle, is equal to the length of the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent.

In triangle OPA and OPB

∠OPA = ∠OPB

OA = OB (radii)

OP = OP (common)

ΔOPA is congruent to ΔOPB by RHS criterion.

We know that if all the corresponding sides and angles of a triangle are equal then the two triangles are congruent.

Therefore, PA = PB

Hence, we proved that the lengths of the tangents drawn from the external point P to a circle are equal.