If a circle 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)
Solution:
Given, a circle touches the side BC of a triangle ABC at P
The circle touches the extended sides AB and AC of the triangle at Q and R.
We have to prove that AQ = 1/2 (BC + CA + AB)
We know that the tangents drawn through an external point to a circle are equal.
So, BP = BQ --------------- (1)
CP = CR -------------------- (2)
AQ = AR -------------------- (3)
We know that the perimeter of a triangle is the sum of all the three sides of a triangle.
Perimeter of triangle ABC = AB + BC + AC
From the figure,
BC = BP + PC
So, AB + BC + AC = AB + (BP + PC) + AC
From (1) and (2),
= AB + BQ + CR + AC
From the figure,
AB + BQ = AQ [Symmetric Property]
CR + AC = AR [Symmetric Property]
As the tangents drawn through an external point to a circle are equal
BQ + CR = BC
So, AB + BC + AC = AQ + AR
From (3),
AB + BC + AC = AQ + AQ
AB + BC + AC = 2AQ
So, AQ = 1/2(AB + BC + AC)
Therefore, it is proven that AQ = 1/2(AB + BC + AC)
✦ Try This: In Δ ABC,AB = AC If the interior circle of ΔABC touches the sides AB, BC and CA at D, E and F respectively. Prove that E bisects BC.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.4 Sample Problem 2
If a circle 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)
Summary:
If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively. It is proven that AQ = 1/2 (BC + CA + AB)
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