2022 AMC 10A Problem 13

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 $\boxed{\textbf{(C)}\ 10}$.

A perpendicular to an angle bisector is the classic signal to reflect: the reflection of $B$ over the bisector lands on the other side of the angle at the same distance from $A$, creating an isosceles triangle and a known segment $EC$. Once the auxiliary point is on $AC$, the parallel line supplies similar triangles, and the answer is a ratio times $BC$.

Coordinates work but are slower: put $B = (0,0)$, $C = (5,0)$, and use the bisector condition to locate $A$ in terms of one parameter; the final $AD$ comes out $10$ regardless of the parameter, confirming the answer does not depend on the triangle's shape.

• Trying to compute AB and AC themselves; they are not determined, and only their ratio 2:3 from the Angle Bisector Theorem is needed.
• Pairing the wrong triangles (for example $EAD$ with $EBC$ in the wrong vertex order) and getting the ratio $\frac{2k}{3k}$, which leads to $AD = \frac{10}{3}$ or $7.5$.