#### Question

From the given figure, prove that:

AP + BQ + CR = BP + CQ + AR

Also show that:

AP + BQ + CR = `1/2`× Perimeter of ΔABC.

#### Solution

Since from B, BQ and BP are the tangents to the circle

Therefore, BQ = BP ………..(i)

Similarly, we can prove that

AP = AR …………..(ii)

and CR = CQ ………(iii)

Adding,

AP + BQ + CR = BP + CQ + AR ………(iv)

Adding AP + BQ + CR to both sides

2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR

2(AP + BQ + CR) = AB + BC + CA

Therefore, AP + BQ + CR = `1/2`× (AB + BC + CA)

AP + BQ + CR = `1/2`× perimeter of triangle ABC

Is there an error in this question or solution?

Solution From the Given Figure, Prove That: Ap + Bq + Cr = Bp + Cq + Ar Also Show That: Ap + Bq + Cr = `1/2`× Perimeter of δAbc. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.