Problem — Angle between two altitudes of an acute triangle

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Let $ABC$ be an acute triangle. Draw the altitude to the side $BC$ (with foot $D$) and the altitude to the side $AC$ (with foot $E$); they meet at the point $H$. Prove that the angle $\widehat{AHB}$ formed by the two altitudes is the supplement of the third angle $\widehat{ACB}$. Then find the value of $\widehat{AHB}$ when $\widehat{ACB} = 70^\circ$.

Solution

Proof. Let $\gamma = \widehat{ACB}$. The two altitudes considered have feet $D$ on $BC$ and $E$ on $AC$; by construction $\widehat{HDC} = 90^\circ \quad\text{(since } AD \perp BC\text{)}, \qquad \widehat{HEC} = 90^\circ \quad\text{(since } BE \perp AC\text{)}.$ Consider the quadrilateral $CDHE$, whose vertices are $C$, the foot $D$, the orthocenter $H$ and the foot $E$. The sum of its interior angles is $360^\circ$: $\widehat{DCE} + \widehat{CDH} + \widehat{DHE} + \widehat{HEC} = 360^\circ.$ Since $\widehat{DCE} = \gamma$ (it is the angle $\widehat{ACB}$) and $\widehat{CDH} = \widehat{HEC} = 90^\circ$, $\gamma + 90^\circ + \widehat{DHE} + 90^\circ = 360^\circ \quad\Longrightarrow\quad \widehat{DHE} = 180^\circ - \gamma.$ The angle $\widehat{AHB}$ is vertical to $\widehat{DHE}$ (the side $HA$ lies on the line $HD$ and the side $HB$ on the line $HE$), hence $\widehat{AHB} = \widehat{DHE} = 180^\circ - \gamma$: the angle between the two altitudes is the supplement of $\widehat{ACB}$. $\;\blacksquare$

Numerical case. With $\gamma = 70^\circ$: $\widehat{AHB} = 180^\circ - 70^\circ = 110^\circ.$ $\boxed{\widehat{AHB} = 110^\circ}$