In triangle $\triangle{ABD}$, $C\in BD$, $E\in AD$, $BE\cap AC =\{F\}$ $B,F$ and $E$ are collinear. $AB$ is the angle bisector of the $\measuredangle{HAC}$. $\measuredangle{HAB}=\measuredangle{BAC}=50^\circ$, $\measuredangle{HDF}=\measuredangle{BDF}=20^\circ$ is given. Find $\measuredangle{CED}=\alpha$.

Following picture is a diagram for visualizing the question: enter image description here Some obvious facts: $\measuredangle{CAD}=80^\circ$ (also marked in diagram) and $\measuredangle{ABC}=10^\circ$, $\measuredangle{ACD}=60^\circ$. Also triangle $\triangle{AFD}$ is an isosceles triangle because $\measuredangle{AFD}=80^\circ$. And lastly if you draw an angle bisector of the angle $\measuredangle{ADF}$, that intersects with $AF$ with a point $I$, $\measuredangle{DIC}=90^\circ$ can be observed. So triangle $\triangle{DCI}$ have angles: $30^\circ-60^\circ-90^\circ$ I don't know whether these are appropriate for solving the problem. Thanks!


I think i have gotten over a little bit. Let $T=DF\cap AB$. $CI$ must be the angle bisector of the $\measuredangle{ACB}$ because $T$ is the intersection point of $DF$ and $AB$ (one of them is the angle bisector of $\measuredangle{ADC}$ and one of them is the angle bisector of $\measuredangle{HAC}$.) So $\measuredangle{ACT}=60^\circ$, $\measuredangle{TFC}=80^\circ$ and $\measuredangle{FTC}=40^\circ$. Notice that $ACD\cong FCT$. enter image description here

  • $\begingroup$ There seem to be some assumptions being made in this diagram that aren't listed in the problem. Namely: $\{B,C,D\}$ are taken to be collinear as are $\{A,D,E,H\}$. Without these, the diagram shown isn't the only one possible. $\endgroup$ – Semiclassical May 27 '17 at 19:02
  • $\begingroup$ I edited the question @Semiclassical $\endgroup$ – user373239 May 28 '17 at 9:19
  • $\begingroup$ A warning to solvers: the angles between $BE$ and any of the other lines are not nice; it seems that they are irrational. $\endgroup$ – Rosie F Jan 21 at 15:32

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