As shown in the figure, it is known that in the triangle ABC, ab = AC, ad is the height on the edge of BC, and point P is in the triangle abd

As shown in the figure, it is known that in the triangle ABC, ab = AC, ad is the height on the edge of BC, and point P is in the triangle abd

Prove: because ABC is isosceles triangle, so D is the midpoint of BC, because P is in abd, so angle BAP is less than angle cap, in triangle BCP, because BP is less than PC, so angle PCB is less than angle PBC, so angle ACP is greater than angle ABP, so angle APB > angle APC