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Properties of Triangles

Section: Mathematics

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Question : 3 of 50

Marks: +1, -0

In a triangle

A B C

(r_1-r_3) (r_1-r_2) -2 r_2 r_3=0

a^2-b^2=

[AP EAPCET 23 May 2025 S2]

$c^2+\;\frac{b^2}{4}$
$c^2$
abc
$\;\frac{b+a}{c}$

Solution:

\because (r_1-r_3) (r_1-r_2) -2 r_2 r_3=0

and we know that

r_1=\;\frac{\Delta}{s-a}, r_2=\;\frac{\Delta}{s-b}, r_3=\;\frac{\Delta}{s-c}

substitute these values

\Rightarrow \Delta^2 \left(\;\frac{1}{s-a}-\;\frac{1}{s-c}\right) \left(\;\frac{1}{s-a}-\;\frac{1}{s-b}\right) -2 \Delta^2 \left(\;\frac{1}{(s-b)(s-c)}\right) =0

\;\Rightarrow \;\; \left(\;\frac{s-c-s+a}{(s-a)(s-c)}\right) \left(\;\frac{s-b-s+a}{(s-a)(s-b)}\right) - \left(\;\frac{2}{(s-b)(s-c)}\right) =0

\Rightarrow \;\frac{1}{(s-b)(s-c)} \left[\;\frac{(a-c)(a-b)}{(s-a)^2}-2\right] =0

\Rightarrow (a-c)(a-b)-2(s-a)^2=0

\Rightarrow (a-c)(a-b)-2 \left(\;\frac{a+b+c}{2}-a\right)^2=0

\Rightarrow 2(a-c)(a-b)-(b+c-a)^2=0

\Rightarrow 2 (a^2-a b-a c+b c) - (b^2+c^2+a^2) -2(b c-c a-a b)=0

\Rightarrow a^2-b^2-c^2-2 a b-2 a c+2 b c-2 b c+2 a c+2 a b=0

\Rightarrow a^2-b^2=c^2

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