Problem — Angle between two internal bisectors of a triangle

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In a triangle $ABC$ the bisectors of the interior angles at vertices $A$ and $B$ are drawn, meeting at the point $I$. Prove that the obtuse angle $\widehat{AIB}$ equals a right angle increased by half of the third angle $\widehat{ACB}$. Then apply the result to the case $\widehat{ACB} = 50^\circ$.

Solution

Proof. Let $\alpha$, $\beta$, $\gamma$ denote the measures of the interior angles at vertices $A$, $B$, $C$; then $\alpha + \beta + \gamma = 180^\circ$.

The bisector at $A$ makes with the side $AB$ an angle $\widehat{IAB} = \dfrac{\alpha}{2}$; the bisector at $B$ makes with $AB$ an angle $\widehat{IBA} = \dfrac{\beta}{2}$. In triangle $AIB$ the sum of the interior angles gives $\widehat{AIB} = 180^\circ - \frac{\alpha}{2} - \frac{\beta}{2} = 180^\circ - \frac{\alpha + \beta}{2}.$ Since $\alpha + \beta = 180^\circ - \gamma$, substituting yields $\widehat{AIB} = 180^\circ - \frac{180^\circ - \gamma}{2} = 180^\circ - 90^\circ + \frac{\gamma}{2} = 90^\circ + \frac{\gamma}{2}.$ The angle $\widehat{AIB}$ is therefore a right angle increased by half of the third angle, and it is obtuse because $\gamma > 0^\circ$. $\;\blacksquare$

Numerical case. With $\gamma = \widehat{ACB} = 50^\circ$: $\widehat{AIB} = 90^\circ + \frac{50^\circ}{2} = 90^\circ + 25^\circ = 115^\circ.$ $\boxed{\widehat{AIB} = 115^\circ}$