A circle with an interior triangle ABC, the bisector of angle A intersecting the circle at D, and an interior center I, and a secondary center P (P is on the bisector of angle A), then DA = DI = DC = DP, the figure is drawn as a chicken's claw. You just got started in geometry right, so post any questions later and I'll take a look! The proof is: set BAI = CAI (represents the angle) = @, ABI = CBI = #, and CBD = CAD = @, then IBD = # + @, BID = BAI + ABI + @ + #, so BD = DI, and similarly, DI = DC, and because the triangle IBP is a right triangle, BD = DI, and therefore DP = DI, get the proof.