Q25 of 49 Page 1

In Fig. 5, the sides AB, BC and CA of triangle ABC touch a circle with centre O and radius r at P, Q and R respectively.

Prove that:


(i) AB + CQ = AC + BQ


(ii) Area (Δ ABC) = 1/2 (Perimeter of ΔABC) × r


(i)


From the figure,


AP = AR [Tangents from an external point to the same circle are equal]


BP = BQ [Tangents from an external point to the same circle are equal]


CQ = CR [Tangents from an external point to the same circle are equal]


Now,


AP + BP + CQ = AR + BQ + CR


⸫ AB + CQ = AC + BQ [⸪ AB = AP + PB and AC = AR + RC]


(iii) Area (ΔABC) = Area (Δ OBC) + Area (ΔOAC) + Area (ΔOAB)


=


=


= × (Perimeter ΔABC) × r


Hence Proved.


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