Question

A circle is circumscribed in a triangle ABC. the sides AB ,BC ,CA of the triangle touches the circle at points P, Q,R respectivelyprove that :i)AB + CQ =AC + BQii)area of triangle ABC =1/2 (perimeter of triangle ABC) *r where r =OQ= the radius of the circle

Answer 1

Let's prove the first part:Using the property, lengths of tangent from an external point are equal in lengths,AP = AR → (1)BP = BQ → (2)CQ = CR → (3)Adding (1) and (2),AB = AR + BQ → (4)Adding (1) and (3),AP + CQ = AC ⇒AC = AP + CQ → (5)Subtracting (4) from (5),AC AB = AP + CQ AR BQSince AP = AR,AC AB = CQ BQAC + BQ = AB + CQLet's prove the second part:ar (ΔABC) = ar (ΔAOB) + ar (ΔBOC) + ar (ΔCOA)Since radius is perpendicular to the tangent at the point of contact,ar (ΔABC) = ar (ΔAOB) + ar (ΔBOC) + ar (ΔCOA) =