MathSolver94 wrote: Let $ABC$ be a triangle in which $AB=AC$. A point $I$ lies inside the triangle such that $\angle ABI=\angle CBI$ and $\angle BAI=\angle CAI$. Prove that \[\angle BIA=90^o+\dfrac{\angle C}{2}\] incentre and $180-B/2-A/2$ as asked? For which age is this?

SCP wrote: MathSolver94 wrote: Let $ABC$ be a triangle in which $AB=AC$. A point $I$ lies inside the triangle such that $\angle ABI=\angle CBI$ and $\angle BAI=\angle CAI$. Prove that \[\angle BIA=90^o+\dfrac{\angle C}{2}\] incentre and $180-B/2-A/2$ as asked? For which age is this? bongsu: 13-14 age It doesnt state that $I$ is an incenter. It is just a point.

MathSolver94 wrote: SCP wrote: MathSolver94 wrote: Let $ABC$ be a triangle in which $AB=AC$. A point $I$ lies inside the triangle such that $\angle ABI=\angle CBI$ and $\angle BAI=\angle CAI$. Prove that \[\angle BIA=90^o+\dfrac{\angle C}{2}\] incentre and $180-B/2-A/2$ as asked? For which age is this? bongsu: 13-14 age It doesnt state that $I$ is an incenter. It is just a point. Isn't that trivial?

What do u mean? Too easy?

Proofs are too advanced for the middle school forums.

MathSolver94 wrote: What do u mean? Too easy? Excuse me? Sorry if I am wrong. But this problem seems direct and the $AB=AC$ is not used. In triangle $AIB$, $\angle ABI=\frac{B}{2}, \angle BAI=\frac{A}{2}$ and so, since sum of angles in a triangle is $180^{\circ}$, $\angle AIB=180^{\circ}-\frac{B}{2}-\frac{A}{2}=\angle BIA=90^{\circ}+\frac{\angle C}{2}$ Did I misunderstand the question?

$I$ is the incenter of $ABC$,hence the result.

By construction, $I$ is the incenter of $\triangle ABC$. Hence, $\angle BIA=180^{\circ}-(\angle IBA+\angle IAB)=180^{\circ}-(\frac{1}{2}\angle A+\frac{1}{2}\angle B)=90^{\circ}+\frac{1}{2}\angle C$. $\blacksquare$

angle ABI= angle CBI=B/2 angle BAI=angle CAI=A/2 angle AIB=180°-(B/2+A/2)=90°+C/2