In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

Question

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

What I Tried: Here is a picture :-

You can see I tried angle-chasing, but that really did not lead me to any useful info. Also I tried Pythagoras theorem, taking $AP = k$ and $PB = (6 – k)$ gives :-
$$k^2 + 9 = AP^2$$
$$(6 – k)^2 + 9 = PC^2$$
And that is $3$ variables in $2$ equations, so I need to find another value before trying this.

I think the fact that $AD = \frac{AB}{2}$ should be used somehow, maybe by some construction or something. Also in Geogebra I saw that $\angle APD = 65.7^\circ$ , $\angle CPB = 32.85^\circ$ , which are $2$ odd-looking angles, and $AP$ came to be $1.35$ .

Can anyone help me how is this the case? Thank You.

Answer 1

Let $a=AP$ be the unknown to be found, we are trying to write an equation that $a$ should satisfy.

Let $Q$ be the point of intersection of the angle bisector of $\hat P$ in $\Delta APD$ with the opposite side $AD$ in this triangle. So $PQ$ divides this angle $\hat P$ in two angles of measure $x$, notation as in the OP. Then we have successively:

• $PD$ is $\sqrt{3^2+a^2}$, Pythagoras.
• $\Delta APQ\sim\Delta BPC$, so we obtain a formula for $AQ$ in terms of $a$, since $AQ:a=BC:(6-a)$. Explicitly, $AQ=3a/(6-a)$.
• This gives also $QD=AD-AQ=3-AQ$ in terms of $a$. Explicitly, $QD=(18-6a)/(6-a)$.
• We write now the angle bisector theorem in $\Delta APD$: $$ \frac{QD}{QA}=\frac {PD}{PA}\ . $$ this is the needed equation. Making things explicit, the resulted equation is: $$ \frac{18-6a}{3a}=\frac{\sqrt{9+a^2}}{a}\ . $$ We simplify with $a>0$, get $6-2a=\sqrt{9+a^2}$. So $6-2a>0$ (i.e. $a<3$), and $$ (6-2a)^2 = 9+a^2\ . $$ This is $a^2-8a+9=0$, the two roots are $4\pm 7$, and we pick the one $<3$, which is $4-\sqrt 7$.