2022 AMC 10A Problem 13

1. Ratio of the sides. The Angle Bisector Theorem says the bisector from $A$ cuts $BC$ in the same ratio as the two sides next to $A$:
$$\frac{AB}{AC} = \frac{BP}{PC} = \frac23.$$
So write $AB = 2k$ and $AC = 3k$ for some number $k$.
2. Extend the perpendicular line. The line through $B$ perpendicular to $AP$ also crosses line $AC$; call that crossing point $E$. We add $E$ because it will link $B$, $D$ and side $AC$.
3. Why $AE = AB$. Let $F$ be where $BE$ crosses $AP$. Triangles $ABF$ and $AEF$ share side $AF$, have equal angles at $A$ (since $AP$ is the bisector), and both have a right angle at $F$. So they are congruent, and $AE = AB = 2k$. You can picture it as folding the triangle along $AP$: point $B$ lands on $E$.
4. Find $EC$. Since $AE = 2k$ is less than $AC = 3k$, point $E$ lies between $A$ and $C$, and $EC = 3k - 2k = k$.
5. Similar triangles. Point $D$ is on line $BE$ and on the line through $A$ parallel to $BC$. Look at triangles $EAD$ and $ECB$:
- the angles at $E$ are equal (vertical angles);
- $AD$ is parallel to $BC$, so the angles at $A$ and $C$ are equal (alternate angles).

So the triangles are similar, with $EA$ matching $EC$ and $AD$ matching $CB$.
6. Use the ratio. $\dfrac{AD}{CB} = \dfrac{EA}{EC} = \dfrac{2k}{k} = 2$. Since $BC = 2 + 3 = 5$, we get $AD = 2 \times 5 = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.

A line perpendicular to an angle bisector is a signal to fold across the bisector. The fold makes an isosceles triangle and carries $B$ over to the other side of the angle, where it creates a length ($EC$) you can compute. After that, the parallel line gives similar triangles, and the answer is just a ratio times $BC$.

By the Angle Bisector Theorem, $\dfrac{AB}{AC} = \dfrac{BP}{PC} = \dfrac{2}{3}$; write $AB = 2k$ and $AC = 3k$.

Let the line through $B$ perpendicular to $AP$ meet line $AC$ at $E$. In triangle $ABE$, the line $AP$ both bisects the angle at $A$ and is perpendicular to the opposite side $BE$, so $ABE$ is isosceles with $AE = AB = 2k$ (equivalently, $E$ is the reflection of $B$ across $AP$). Since $AE = 2k < 3k = AC$, $E$ lies between $A$ and $C$, and $EC = k$.

Now $D$ lies on line $BE$ and on the line through $A$ parallel to $BC$. Lines $BD$ and $AC$ cross at $E$, cutting the two parallel lines $AD$ and $BC$, so triangles $EAD$ and $ECB$ are similar (vertical angles at $E$, alternate interior angles at $A$ and $C$). Hence
$$\frac{AD}{CB} = \frac{EA}{EC} = \frac{2k}{k} = 2, \qquad AD = 2 \cdot BC = 2 \cdot 5 = 10.$$

The answer is $\boxed{\textbf{(C)}\ 10}$.