MathLabs

Problem 2

Let P P be a point inside triangle ABC ABC such that ∠APB−∠ACB=∠APC−∠ABC\angle APB-\angle ACB=\angle APC-\angle ABC . Let D D and E E be the incenters of triangles APB APB and APC APC , respectively. Show that AP AP , BD BD , and CE CE meet at a point.
Step 4 of 4: Apply two angle-bisector theorems
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In plain words

Apply two angle-bisector theorems

AWWP=ABPB=ACPC\dfrac{AW}{WP}=\dfrac{AB}{PB}=\dfrac{AC}{PC}
Detailed analysis

Let W=BD∩AP W=BD\cap AP . Since D D is the incenter of APB APB , BD BD bisects angle ABP ABP , so the angle-bisector theorem gives AW/WP=AB/PB AW/WP=AB/PB . The equality above then gives AW/WP=AC/PC AW/WP=AC/PC , which is exactly the angle-bisector condition in triangle APC APC ; hence W W lies on CE CE , and AP,BD,CE AP,BD,CE are concurrent.