## Online Course Discussion Forum

### Math II-A Geometry 1.28

Math II-A Geometry 1.28

Let ABC be a triangle with AB = AC and ∠BAC = 20°, and let P be a point on side AB such that AP = BC. Find ∠ACP. I know that ABC is a isosceles triangle with B and C as 80° each. Since two segments are equal to each other, there is probably something involving either congruent or similar triangles, or both.  Can I get a hint for how to solve this? Thank you.

Re: Math II-A Geometry 1.28

You are on the right track.  Since $AP=BC$, can you make a new triangle with base $AP$ and congruent to $\triangle ABC$?