Mathematics - Properties of Triangle Question with Solution | TestHub

The condition $r_a{r}_b=r{r}_c$ implies $\frac{{\Delta }^2}{(s-a)(s-b)}=\frac{{\Delta }^2}{s(s-c)}$, which simplifies to $s(s-c)=(s-a)(s-b)$. Expanding this gives $s^2-sc=s^2-s(a+b)+ab$, so $s(a+b-c)=ab$. Using $s=(a+b+c)/2$, this means $(a+b+c)(a+b-c)=2ab$, or $(a+b{)}^2-c^2=2ab$. This simplifies to $a^2+b^2-c^2=0$, so $a^2+b^2=c^2$. Thus, the triangle is right-angled at $C$. For a right triangle at $C$: A) $r+R=(a+b-c)/2+c/2=(a+b)/2$. (Correct) B) $r_a+r_b=(s-b)+(s-a)=(a+c-b)/2+(b+c-a)/2=c$. (Correct) C) $2R=c$. Also $r_a+r_b-r_c+r=c-s+(s-c)=c$. (Correct) D) $\sin A+\sin B=\sin A+\cos A$, which is not generally $\sqrt{2}$.