CBSE2026Class Class 10 · Mathematics3 Marks · Short✅ Verified
(a) A circle touches the side BC of a ΔABC at P and also touches the sides AB and AC produced at Q and R respectively. Prove that AR = (1/2)(perimeter of ΔABC).
OR
(b) Prove that the tangents drawn to a circle at the end points of a diameter are parallel to each other.
✅ Answer & Solution
**(a)** Tangents from an external point are equal in length.
From A: $AQ=AR$
From B: $BQ=BP$
From C: $CR=CP$
$AR=AQ=AB+BQ=AB+BP$
$AR=AC+CR=AC+CP$
Adding: $2AR=AB+AC+(BP+CP)=AB+AC+BC=\text{Perimeter of }\triangle ABC$
$$\therefore AR=\frac12(\text{Perimeter of }\triangle ABC)$$
**(b)** Let AB be a diameter of a circle with tangents at A and B. Since the tangent at any point is perpendicular to the radius at that point, tangent at A $\perp$ OA and tangent at B $\perp$ OB. Since O, A, B are collinear (AB is a diameter), both tangents are perpendicular to the same line AB, hence they are parallel to each other.
