Similar triangles eventually

Question

Through the A-peak of the ABCD square is built a right that crosses the country BC at the point P. The bisector(bisectrix) of the angle PAD crosses the side to the point L. Find DL + BP if AP = 5 cm

Answer 1

Let the side of the square be $s = AD = AB$ and let $\theta = \angle DAL = \angle LAP$. Also let lengths $x = AP, y = DL$ and $z = BP$.

Then $\angle ABP = 90^\circ - 2\theta$. In the right triangle $ABP$ we have $$ z = x \sin (90^\circ - 2\theta) = x \cos(2\theta)= x (1 - 2\sin^2\theta) \\ s = x \cos(90^\circ - 2\theta) = x \sin(2\theta)= 2x \sin \theta \cos \theta $$ And in right triangle $ADL$ we have $$y = s\tan \theta = (2x\sin\theta \cos \theta) \frac{\sin\theta}{\cos\theta} = 2x \sin^2 \theta $$ Adding, $$ z+y = x(1 - 2\sin^2\theta)+2x\sin^2\theta=x $$ In the problem given, $x = 5$ cm. so $AL + AP= z+y=x = 5$ cm.

Answer 2

hint

Let $g$ be the angle $(\vec {AB},\vec {AP}) $ and $a=AB $. then

$$\sin(g)=\frac {BP}{5} $$ and $$\tan (g)=\frac {BP}{a} $$ on the other hand

$$\tan (\frac {90-g}{2})=\frac {DL}{a} $$