MP Board · Class 10 · Mathematics · CirclesProve that the lengths of tangents drawn from an external point to a circle are equal. Using this theorem, find the perimeter of a triangle $ABC$ circumscribing a circle of radius $4\text{ cm}$, if the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8\text{ cm}$ and $6\text{ cm}$ respectively.

Proof of Theorem:

Given: A circle with center $O$, an external point $P$, and two tangents $PQ$ and $PR$ drawn from $P$ to the circle touching at points $Q$ and $R$ respectively. To Prove: $PQ = PR$ Construction: Join $OP$, $OQ$, and $OR$.

Proof:

1. Radius is perpendicular to the tangent at the point of contact. Therefore, $\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$.
2. In right-angled triangles $\triangle OQP$ and $\triangle ORP$:
   • $OQ = OR$ (Radii of the same circle)
   • $OP = OP$ (Common hypotenuse)
   • $\angle OQP = \angle ORP = 90^\circ$
3. By RHS congruence criterion, $\triangle OQP \cong \triangle ORP$.
4. Therefore, $PQ = PR$ (by CPCTC).

Numerical Solution:

Given:

• Radius of the circle, $OD = OF = OE = 4\text{ cm}$ ($E$ and $F$ are contact points on $AB$ and $AC$ respectively).
• $BD = 8\text{ cm} \implies BE = 8\text{ cm}$ (tangents from $B$)
• $DC = 6\text{ cm} \implies FC = 6\text{ cm}$ (tangents from $C$)
• Let $AE = AF = x$ (tangents from $A$)

Sides of the Triangle:

• $AB = x + 8$
• $BC = BD + DC = 8 + 6 = 14\text{ cm}$
• $AC = x + 6$

Semi-perimeter ($s$): $$s = \frac{AB + BC + AC}{2} = \frac{(x + 8) + 14 + (x + 6)}{2} = \frac{2x + 28}{2} = x + 14$$

Area of $\triangle ABC$ using Heron's Formula:

• Area $= \sqrt{s(s-a)(s-b)(s-c)}$
• $s - AB = (x + 14) - (x + 8) = 6$
• $s - BC = (x + 14) - 14 = x$
• $s - AC = (x + 14) - (x + 6) = 8$
• $\text{Area} = \sqrt{(x + 14)(6)(x)(8)} = \sqrt{48x(x + 14)}$

Area of $\triangle ABC$ as sum of three triangles ($OAB, OBC, OCA$):

• $\text{Area} = \text{Area}(OAB) + \text{Area}(OBC) + \text{Area}(OCA)$
• $\text{Area} = \frac{1}{2} \times AB \times OD + \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AC \times OE$
• $\text{Area} = \frac{1}{2} \times (x + 8) \times 4 + \frac{1}{2} \times 14 \times 4 + \frac{1}{2} \times (x + 6) \times 4$
• $\text{Area} = 2(x + 8 + 14 + x + 6) = 2(2x + 28) = 4x + 56$

Equating both areas: $$\sqrt{48x(x + 14)} = 4(x + 14)$$\nSquaring both sides: $$48x(x + 14) = 16(x + 14)^2$$\nDividing both sides by $16(x + 14)$ [since $x + 14 \neq 0$]: $$3x = x + 14 \implies 2x = 14 \implies x = 7\text{ cm}$|

Perimeter of $\triangle ABC$: $$\text{Perimeter} = 2s = 2(x + 14) = 2(7 + 14) = 2(21) = 42\text{ cm}$$