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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