Question

please someone solve this problem....... ​

Answer 1

Answer:

Hence proved...

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

Answer 2

Answer:

Step-by-step explanation:

Since tangents from an exterior point to a circle are equal in length.

Therefore, BP = BQ........[From B]... (1)

CP = CR.......[From C]... (2)

AQ = AR..... [From A]....(3)

From(3),we have

AQ = AR

=>. AB + BQ = AC + CR

=>. AB + BP = AC + CO.....

[From 1 and 2] ......(4)

Now,

Perimeter of triangle ABC=

BC + CA + AB..... (5)

Perimeter of triangle ABC=

AB + BC + AC

Therefore,

BC + CA + AB = AB + BC + AC

BC + CA + AB = AB +(BP + PC) + AC

BC + CA + AB=(AB + BP)+ (AC + PC)

BC + CA + AB = 2(AB + BP)

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

Therefore,

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