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Question
If ΔABC is isosceles with AB = AC and C (0, 2) is the in circle of the ΔABC touching BC at L, prove that L, bisects BC.
Solution
Given ΔABC is isosceles AB = AC
We know that
The tangents from external point to circle are equal in length
From point A, AP = AQ
But AB = AC ⇒ AP + PB = AQ + QC
⇒ PB = PC …. (i)
From B, PB = BL; ….(ii) from C, CL = CQ …..(iii)
From (i), (ii) & (iii)
BL = CL
∴ L bisects BC.
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