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 $CE$
Best Answer
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.