In triangle ABC, AF is the altitude to BC and AD is the bisector of angle BAC. Given angle B = 36 degrees and angle C = 76 degrees, find the three angles of triangle ADF.

Step-by-step solution

1. 1

   Find the apex angle. The angles of $\triangle ABC$ sum to $180^{\circ}$:

   $\angle A=180^{\circ}-36^{\circ}-76^{\circ}=68^{\circ}$

2. 2

   Use the altitude to get ∠BAF. In right triangle $ABF$ the angle at $F$ is $90^{\circ}$, so

   $\angle BAF=90^{\circ}-\angle B=90^{\circ}-36^{\circ}=54^{\circ}$

3. 3

   Use the bisector to get ∠BAD.

   $\angle BAD=\frac{\angle A}{2}=\frac{68^{\circ}}{2}=34^{\circ}$

   Since $34^{\circ}<54^{\circ}$, the ray $AD$ lies between $AB$ and $AF$, so $D$ sits between $B$ and $F$ on the base.

4. 4

   Subtract to get the angle at A in triangle ADF.

   $\angle DAF=\angle BAF-\angle BAD=54^{\circ}-34^{\circ}=20^{\circ}$

5. 5

   Read off the other two angles. $AF\perp BC$ and $D$ lies on $BC$, so

   $\angle AFD=90^{\circ}$

   and the angles of $\triangle ADF$ must sum to $180^{\circ}$:

   $\angle ADF=180^{\circ}-20^{\circ}-90^{\circ}=70^{\circ}$

   $\boxed{\angle DAF=20^{\circ},\ \angle AFD=90^{\circ},\ \angle ADF=70^{\circ}}$

6. 6

   Verify ∠ADF a second way, through triangle ABD. In $\triangle ABD$: $\angle ABD=36^{\circ}$ and $\angle BAD=34^{\circ}$, so $\angle ADB=110^{\circ}$. Since $B$, $D$, $F$ are collinear, $\angle ADF=180^{\circ}-110^{\circ}=70^{\circ}$ ✓ — matching the first route.

7. 7

   Note the general rule this illustrates. The angle between the altitude and the bisector from the same vertex is always half the difference of the other two angles: $\tfrac12\left(\angle C-\angle B\right)=\tfrac12(76^{\circ}-36^{\circ})=20^{\circ}$, exactly $\angle DAF$. Beware of worked solutions that report $\angle AFD$ as anything other than $90^{\circ}$ — the foot of an altitude forces a right angle there.