Question

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

• A. isosceles
• B. equilateral
• C. right angled but not isosceles
• D. scalene triangle

Answer 1

The correct option is A isosceles

We have,
$c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\therefore c(a+b)\sqrt{\frac{s(s-b)}{ca}}=b(a+c)\sqrt{\frac{s(s-c)}{ab}}\Rightarrow (a+b)\sqrt{c(s-b)}=(a+c)\sqrt{b(s-c)}$
squaring both sides,
$(a+b{)}^{2}c(s-b)=(a+c{)}^{2}b(s-c)\Rightarrow s\left[c\right(a+b{)}^2-b(a+c{)}^2]-bc\left[\right(a+b^2)-(a+c{)}^2]=0\Rightarrow s[c{a}^2+2abc+c{b}^2-b{a}^2-2abc-b{c}^2]-bc(b-c\left)\right(2a+b+c)=0\Rightarrow s[bc(b-c)-a^2(b-c)]-bc(b-c\left)\right(2a+b+c)=0\Rightarrow (b-c\left)\right[s(bc-a^2)-bc(2a+b+c)]=0\Rightarrow (b-c\left)\right[s(bc-a^2)-bc(2s+a)]=0\Rightarrow -(b-c\left)\right[s(bc+a^2)+abc]=0$
since, $a,b,c$ are all positive and so $s(bc+a^2)+abc\ne 0$
$\Rightarrow b-c=0\Rightarrow b=c$
So, $△ABC$ is isosceles.