If $m\angle B = 60^\circ$ and $BD$ bisects that angle, what are $m\angle ABD$ and $m\angle CBD$? How does that make the two sides $\overline{BA}$ and $\overline{BC}$ look relative to $BD$?

When a line from a vertex both bisects the angle and is perpendicular to the opposite side, the triangle is symmetric across that line. What does that say about $AB$ and $BC$, and therefore about $\angle A$ and $\angle C$?

Once you know that $\angle A$ and $\angle C$ are equal, use $m\angle A + m\angle B + m\angle C = 180^\circ$ with $m\angle B = 60^\circ$ to solve for the equal base angles.

Set up points to reflect the given conditions. In Desmos, define:

• B = (0,0)
• k = 1 (or any positive number)
• Since $BD$ is an altitude along the $x$-axis, let D = (k,0)
• Because $BD$ bisects the $60^\circ$ angle at $B$, place:
  • A = (k, k*tan(30))
  • C = (k, -k*tan(30)) This makes $\angle ABD = \angle CBD = 30^\circ$, so $\angle ABC = 60^\circ$, and $BD$ is perpendicular to $AC$.

To check that $AB = BC$, compute the squared lengths (to avoid square roots):

• AB2 = (k-0)^2 + (k*tan(30)-0)^2
• BC2 = (k-0)^2 + (-k*tan(30)-0)^2 Desmos will show that AB2 and BC2 have the same numerical value, confirming that $AB = BC$ and the base angles at $A$ and $C$ are equal.

Since $\angle A$ and $\angle C$ are equal and $\angle B$ is $60^\circ$, the two equal angles together must sum to $180^\circ - 60^\circ$. In Desmos, type the expression (180 - 60)/2 to see the measure each of those base angles must have. That value is the measure of $\angle C$.

We are told that in $\triangle ABC$, point $D$ lies on $\overline{AC}$ and $\overline{BD}$ has two special properties:

• Altitude to $\overline{AC}$ means $BD$ is perpendicular to $AC$, so $\angle BDA$ and $\angle BDC$ are right angles: $90^\circ$.
• Bisector of $\angle ABC$ means $BD$ splits $\angle B$ into two equal angles, so

So $BD$ is both perpendicular to $AC$ and exactly down the middle of $\angle B$. This suggests the triangle is symmetric across $BD$.

Because $BD$ is the angle bisector of $\angle B$ and also an altitude to the opposite side $AC$, the triangle is symmetric across line $BD$.

That symmetry means:

• Side $\overline{BA}$ is the mirror image of side $\overline{BC}$ across $BD$, so $AB = BC$.
• In a triangle with $AB = BC$, the base angles at $A$ and $C$ are equal:

Thus, $\triangle ABC$ is isosceles with equal angles at $A$ and $C$.

In any triangle, the sum of the three interior angles is $180^\circ$:

We know $m\angle B = 60^\circ$ and we found $m\angle A = m\angle C$. Let $x = m\angle C$. Then $m\angle A = x$ as well, so

So $m\angle C = 60^\circ$, which matches answer choice C.