Question

In a ∆ABC prove that

$4(\frac{s}{a} -1 ) (\frac{s}{b} -1 ) ( \frac{s}{c}-1) = \frac{r}{R}$​

Answer 1

$\huge{\underline{\mathfrak{\blue{Answer}}}}$

Formulas

$\implies$∆²=s(s-a)(s-b)(s-c)

$\implies r = \dfrac{\triangle}{s}$

SoluTion

$4 \bigg(\dfrac{s-a}{a} \bigg) \bigg(\dfrac{s-b}{b} \bigg) \bigg( \dfrac{s-c}{c} \bigg)$

Multiplying S in numerator and denominator

$\dfrac{4 s(s-a)(s-b)(s-c)}{s\: abc}$

$\implies\dfrac{ \cancel{4} \cancel{\triangle} ^2}{s \cance{4} R \cancel{\triangle}}$

$\implies \dfrac{ \triangle}{ sR} \\ \\ \implies \boxed{\dfrac{r}{R} }$

Answer 2

Question

In a ∆ABC prove that

$4(\frac{s}{a} -1 ) (\frac{s}{b} -1 ) ( \frac{s}{c}-1) = \frac{r}{R}$

Answer

refer to the attachment

hope it helps you