Question

If in $△ABC,c(a+b)cos\frac{1}{2}B=b(a+c)cos\frac{1}{2}C$, prove that the triangle is isosceles.

Answer 1

We have $c(a+b)\cos \frac{1}{2}B=b(a+c)\cos \frac{1}{2}C$
$\therefore c(a+b)\sqrt{\frac{s(s-b)}{ca}}=b(a+c)\sqrt{\frac{s(s-c)}{ab}}$
or $(a+b)\sqrt{c(s-b)}=(a+c)\sqrt{b(s-c)}$
Squarring,$(a+b{)}^{2}c(s-b)=(a+c{)}^{2}b(s-c)$
or $s\left[c\right(a+b{)}^2-b(a+c{)}^2]-bc\left[\right(a+b{)}^2-(a+c{)}^2]=0$
or $s[c{a}^2+2abc+c{b}^2-b{a}^2-2abc-b{c}^2]-bc(2a+b+c)(b-c)=0$
or $s\left[bc\right(b-c)-a^2(b-c\left)\right]-bc(b-c)(2a+b+c)=0$
or $(b-c)\left[s\right(bc-a^2)-bc(2a+b+c\left)\right]=0$
or $(b-c)\left[s\right(bc-a^2)-bc(2s+a\left)\right]=0$
or $-(b-c)\left[s\right(bc+a^2)+abc]=0$
Since a, b, c are all positive and so
$s(bc+a^2)+abc\ne 0$
It follows that $b-c=0$ or $b=c$
Hence $△ABC$ is isosceles.