I have this problem:
Let be given a right triangle $ABC$ $(AB<AC)$; $AH$ is altitude of the triangle, $BD$ is bisector of the angle $\angle ABC$. Let $E$ is projection of the point $C$ onto the line $BD$ and $M$ is intersection of point of two lines $AB$ and $CE$. Prove that $BA\cdot BM + CE\cdot CM = BC^2$.
I tried. We can see $BC=BM$ and $CE=EM$. From $BA\cdot BM + CE\cdot CM = BC^2$, we have $$\dfrac{BA\cdot BM}{BC^2} + \dfrac{CE\cdot CM}{BC^2}=1,$$ then $$\dfrac{BA}{BC} + 2\left( \dfrac{CE}{BC}\right) ^2 = 1.$$ We have $\dfrac{BA}{BC} = \sin \angle BCA $ and $$2\left( \dfrac{CE}{BC}\right) ^2 = 2\sin^2 \angle CBE = 2 \left (\dfrac{1 - \cos\angle ABC}{2}\right) = 1- \cos\angle {ABC} = 1 - \sin BCA. $$ Thus $$\dfrac{BA}{BC} + 2\left( \dfrac{CE}{BC}\right) ^2 =\sin \angle {BCA} + 1 - \sin BCA = 1.$$
Another way, We have $$BA\cdot BM = BA\cdot (BA+AM) = BA^2+BA\cdot AM = BA^2+ AD\cdot AC,$$ and $$ CE\cdot CM = CD\cdot CA=(AC-AD)\cdot CA = AC^2 - AD\cdot CA. $$ Then $$ BA\cdot BM + CE\cdot CM = BA^2+ + AC^2 = BC^2. $$