Question

If the circles touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R respectively, prove that. AQ =1/2 (BC +CA +AB)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given:-

• A circle which touches sides BC of ∆ABC at P and extended sides AB and AC at Q and R respectively.

To prove:-

AQ = 1/2 ( BC + CA + AB )

Proof:-

Tangents drawn from an external point to the circle are equal in length.

AQ = AR ------------------ (1)

BQ = BP ------------------(2)

CR = CP ------------------(3)

Perimeter of ∆ABC = AB + BC + CA

→ AB + BP + PC + CA

→ (AB + BQ) + (CR + CA)

→ AQ + AR

→ AQ + AQ

→ 2AQ

1/2 Perimeter of ∆ABC = AQ

1/2 ( BC + CA + AB)

Hence proved.