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A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.

I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ is the tangent to the circumcircle of $\triangle{ACE}$ as shown, but I am not getting how to prove them. Pure geometrical method is preferred.

enter image description here

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  • $\begingroup$ Can you provide a diagram? $\endgroup$ – SS_C4 Feb 23 '16 at 10:56
  • $\begingroup$ @SS_C4 I don't know how to add a diagram in a question $\endgroup$ – Satvik Mashkaria Feb 23 '16 at 10:58
  • $\begingroup$ meta.math.stackexchange.com/questions/21341/… $\endgroup$ – SS_C4 Feb 23 '16 at 11:00
  • $\begingroup$ And make the image on your computer.(Using something like Paint) $\endgroup$ – SS_C4 Feb 23 '16 at 11:00
  • $\begingroup$ @SatvikMashkaria Shall I try to improve image? $\endgroup$ – Narasimham Feb 23 '16 at 11:45
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enter image description here

Let $F$ be the intersection of line $AD$ and line $BC$. Since $\angle BDA = \angle ABC$, $AF=FB$.

Let $O$ be the center of the circumcircle of $\triangle ABC$ and let $G$, $I$ be the projections of $O$ onto $AF$ and $CD$. Then

  1. $OG = OM$ because $O$ and $F$ lie on the perpendicular bisector of $AB$.
  2. $OG = OI$ because $O$ and $D$ lie on the perpendicular bisector of $AC$.

It follows that $OM=OI$ and that $CO$ is the angle bisector of $\angle DCB$.

Next, consider $\triangle ABF$ with $D$, $M$, and $E$ colinear. Menelaus' theorem implies that $$\frac{EA}{EB}\cdot\frac{MB}{MF}\cdot\frac{DF}{DA}=1.$$

Similarly, $\triangle ABF$ with $H$, $C$, and $E$ colinear. Menelaus' theorem gives us $$\frac{EA}{EB}\cdot\frac{CB}{CF}\cdot\frac{HF}{HA} = 1.$$

Thus we have $$\frac{MB}{MF}\cdot\frac{DF}{DA} = \frac{CB}{CF}\cdot\frac{HF}{HA},$$ or $$\begin{aligned}\frac{HF}{HA} &= \frac{\color{blue}{MB}}{MF}\cdot\frac{DF}{DA}\cdot\frac{CF}{\color{blue}{CB}}\\ &=\frac{DF}{DA}\cdot\frac{CF}{2MF} \end{aligned}$$ Since $AD = CD$ and $2MF = DF+CD+CF$, we have $$\frac{HF}{HA}=\frac{DF\cdot CF}{CD\cdot(DF+CD+CF)}.$$ So $$\frac{HF}{HF+HA}=\frac{DF\cdot CF}{CD(DF+CD+CF)+DF\cdot CF},$$ or $$\frac{HF}{HA} = \frac{DF\cdot CF}{(CD+DF)(CD+CF)}.$$ Since $HA = DA+DF = CD+DF$, we have $$HF = \frac{DF\cdot CF}{CD+DF}.$$ So $CH$ is the angle bisector of $\angle CDF$. Thus $CO\perp CH$, equivalently $EC$ is tangent to the circumcircle of $\triangle ABC$ at $C$.

It follows that $\angle BCE = \angle BAC$, and so $\angle DAC=\angle BEC$.

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