**Q) Prove that the lengths of tangents drawn from an external point to a circle are equal.**

**Using above result, find the length BC of Δ ABC. Given that, a circle is inscribed in Δ ABC touching the sides AB, BC and**

**CA at R, P and Q respectively and AB= 10 cm, AQ= 7cm ,CQ= 5cm**

**Ans: **

**(i) Tangent equal from an external point:**

Let’s connect AO, RO and QO. and compare Δ AOR and Δ AOQ

Here, ∠ ARO = ∠ AQO = 90 (tangent is perpendicular to radius)

OR = OQ (radii of same circle)

OA = OA (common side)

∴ Δ AOR Δ AOQ (by RHS Congruence rule)

∴ AR = AQ (BY CPCT)

**Hence Proved !**

**(ii) Length of BC:**

**Step 1:** Let’s start from point A:

Since AR and AQ are tangents on the circle from external point A,

∴ AR = AQ

Since AQ = 7 cm (given)

∴ AR = AQ = 7 cm

**Step 2:** Now AB = 10 cm (given)

∴ BR = AB – AR = 10 – 7 = 3 cm

**Step 3:** Now BR and BP are tangents from point B on the circle,

∴ BP = BR

**∴ BP = 3 cm** …. (i)

**Step 4:** Next, CQ = 5 cm

Now CQ and CP are tangents from point C on the circle,

∴ CP = CQ

**∴ CP = 5 cm** …. (ii)

**Step 5:** Now BC = BP + CP

∴ BC = 3 + 5 [from equation (i) and (ii)]

∴ BC = 8 cm

**Therefore the length of side BC is 8 cm.**

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