BMO1 B 2020 5

I sort of don’t have the time right now so I can’t provide a diagram but hopefully, my words can explain the solution well:

First, set \angle DPA = x. This means that \angle DAP = x as well as \triangle ADP is an isosceles triangle. Now, because the circumcircle of \triangle ADP is tangent to AC at point A, we can get that \angle CAD = \angle DPA = x.

Now, because quadrilateral ABCD is cyclic, \angle ACD = \angle ABD. From here we can get that \triangle BPD \approx \triangle CAD by AA similarity. But, we can actually make a more powerful statement and say that \triangle BPD \cong \triangle CAD since AD=DP. From here we can derive that BP=AC from congruent triangle properties as the angles facing those sides in the triangles, \angle BDP and \angle CDA are equal.

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