2025 AMC 10A Problem 23

Let the bisector of $\angle B$ meet $AC$ at $D;$ the Angle Bisector Theorem gives $AD = 48$ and $CD = 27$

Show $BD = 48,$ so $\triangle ADB$ is isosceles; then angle-chase to find $\triangle CDP$ isosceles with $PD = CD$

Let the bisector of $\angle B$ hit $AC$ at $D.$ By the Angle Bisector Theorem, $\frac{AD}{DC} = \frac{AB}{BC} = \frac{80}{45},$ and since $AC = 75,$ we get $AD = 48$ and $CD = 27.$ Triangles $BCD$ and $ACB$ share $\angle C,$ with adjacent sides in the common ratio $\tfrac{45}{75}=\tfrac{27}{45}=\tfrac35,$ so they are similar by SAS. Hence $BD=\tfrac35\cdot80=48=AD,$ making $\triangle ADB$ isosceles. Put $\angle DAB=\angle DBA=\theta.$ Because the altitude through $C$ is perpendicular to $AB,$ both $\angle DPC$ and $\angle DCP$ equal $90^\circ-\theta.$ Thus $\triangle CDP$ is isosceles, so $PD=CD=27.$ Since $P$ lies on $BD,$ we get $BP=BD-PD$ $=48-27=21.$ Thus, D is the correct answer.