I'm kinda lost trying to solve this problem:
In the figure, if $\overline {AC} \cong \overline {CB}$, and $\overline {DC} \cong \overline {CE}$. Prove that $\overline {DH} \bot \overline {AB}$
What I know from the figure and the conditions:
Triangle $DCE$ and $ABC$ are isosceles. That the angle $\angle DEC$ is the same as $\angle AEH$. But I don't know where to go from there. My intake of the problem is to prove that the triangle $EAH$ is a right triangle but keep being stuck in the same way. I'm looking for hints more than the solution to the problem =) Thanks!
Best Answer
Let $\angle CAB = \angle CBA = x$ and $\angle CDE = \angle CED = y$. Then, $\angle DCE = 2x$ by exterior angle of triangle. Using angle sum of triangle on $\Delta CDE$, we obtain $2x+2y=180^\circ$, whence $x+y=90^\circ$. Therefore, $\angle HDB + \angle HBD = 90^\circ$. Using angle sum of triangle on $\Delta BDH$, we obtain $\angle DHB=90^\circ$, whence $DH \perp AB$.