Find the length of $CE$

Question

In quadrilateral $ABCD$, $AB=6$, $\angle{ABC}=90°$, $\angle{BCD}=45°$ and $\angle{CAD}=2\angle{ACB}$. If $DE$ is perpendicular to $AC$ with $E$ on side $BC$, Find the length of $CE$.

Could someone give me a hint? Since I can't find any equal side I can't using Law of Sines to find angle of the triangle so it's hard to find $CE$

Picture of the problem

Answer 1

We start from how the figure is drawn.

(1) Put C at (0, 0). (2) Draw the line L: y = –x such that L inclines $45^0$ to the negative x-axis. (3) Draw the line H: y = 6 such that it cuts the positive y-axis at Z. (4) Reflect C about H such that C’ is at (0, 12).

(5) Let A be any point on H and in the 3rd quadrant. (6) Let B be the projection of A on the x-axis. After Joining AC and AC’, we have (i) $\angle BCA = \angle CAZ = \angle ZAC’$; (ii) AZ is the perpendicular bisector of CC’.

(7) The point of intersection of L and AC’ is D which meets the requirement of the question. (8) Let CD cuts H at Y. Then $\angle CYZ = \angle ZYC’ = 45^0$.

(9) Let the perpendicular from D cut AC at M. DM produced will cut BC at E.

[Note: It does not matter where A is. As long as it is on the negative part of H and on the far left of M will meet the given requirement.]

The question is:- "will EYC’ be collinear?". If yes, then CE = 12 because $\triangle ECC'$ is then right and isosceles.

Answer 2

Let $AB=a$, $BC=b$.

Following Mick, let $A=(-b, a), B=(-b, 0)$, $C=(0, 0)$, $C'=(0, 2a)$. Then line $CD$'s equation is $y=-x$ and line $AC'$'s equation is $y=2a+\frac{ax}{b}$. These lines intersect at $D$.

Let $P$ be the foot of the perpendicular from $D$ to $BC$.

Then $D$ is $(-y, y)$ where $y=CP=DP$ and

\begin{align*} y &= 2a-\frac{ay}{b}\\ \implies by &= 2ab-ay\\ \implies (a+b)y &= 2ab\\ \implies y &= \frac{2ab}{a+b} \end{align*}

Then because $DE\perp AC$, $\angle EDP=\angle ACB$, so right-angled triangles $ABC$ and $EPD$ are similar, so \begin{align*} \frac{EP}{DP} &= \frac{AB}{BC}\\ &= \frac{a}{b}\\ \implies EP &= \frac{a\cdot DP}{b}\\ &= \frac{a\cdot2ab}{b(a+b)}\\ &= \frac{2a^2}{a+b}\\ \implies CE&=CP+EP\\ &=\frac{2ab}{a+b}+\frac{2a^2}{a+b}\\ &=\frac{2a(a+b)}{a+b}\\ &=2a=12. \end{align*}