2022 AMC 10A Problem 13

Problem 13 of 25IntermediateGeometry

Let

\triangle ABC

be a scalene triangle. Point

P

lies on

\overline{BC}

so that

\overline{AP}

bisects

\angle BAC.

The line through

B

perpendicular to

\overline{AP}

intersects the line through

A

parallel to

\overline{BC}

at point

D.

Suppose

BP = 2

and

PC = 3.

What is

AD?

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Solution

Consider the following diagram:

Let

Y

be the intersection of

\overline{BD}

and

\overline{AC}.

By the Angle Bisector Theorem,

AB:AC=BP:PC=2:3,

so write

AB=2x

and

AC=3x.

Reflection across the angle bisector

\overline{AP}

sends ray

AB

to ray

AC.

Because

BY\perp AP,

it sends

B

to

Y.

Thus

AY=AB=2x,

and hence

YC=AC-AY=x.

Since

AD\parallel BC,

with

B,Y,D

collinear and

A,Y,C

collinear, we have

\triangle BYC\sim\triangle DYA.

Therefore

\frac{AD}{BC}=\frac{AY}{YC}=2.

Finally,

BC=BP+PC=5,

so

AD=2BC=10.

Thus, C is the correct answer.

More practice

Concepts: angle bisector theorem · similarity · isosceles triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.