Let D be the perpendicular foot relative to BC in a triangle ABC. Let P be the intersection of the internal angular bisector in C with the side AB, and suppose $C\hat{P}A=45^\circ.$ Find $P\hat{D}B$.
I made a simulation in geogebra, and I found the answer, but I could not find a reason using geometric properties. Thanks in advance.
Let $DK$ be a bisector of $\angle ADC$, which interests $PC$ in the point $M$.
Thus, $AM$ is a bisector of $\angle DAC$, which gives $$\measuredangle DAM=\frac{1}{2}\measuredangle DAC=45^{\circ}-\frac{1}{2}\measuredangle ACD.$$ Now, since $\measuredangle ADM=APM=45^{\circ},$ we see that $APDM$ is cyclic.
Thus, $\measuredangle DPM=\measuredangle DAM=45^{\circ}-\frac{1}{2}\measuredangle ACD$.
Id est, $$\measuredangle PDB=\measuredangle DPC+\measuredangle PCD=45^{\circ}-\frac{1}{2}\measuredangle ACD+\frac{1}{2}\measuredangle ACD=45^{\circ}.$$ Done!