2022 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 13 Easy

Let  △ABC  be a scalene triangle. Point P lies on \overline{BC} so that \overline{AP} bisects  ∠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?

  • A.

    8

  • B.

    9

  • C.

    10

  • D.

    11

  • E.

    12

Answer:C

\because AP \perp BD

\therefore AM=AB

\therefore \frac{AB}{AC}=\frac{BP}{PC}=\frac 23

\therefore \frac{MC}{AM}=\frac{AC-AM}{AM}=\frac 12

\because AD \parallel BC

\therefore \triangle {AMD} \sim \triangle {CMB}

\therefore \frac{AD}{BC}=\frac{AM}{MC}=2

\therefore AD=10

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