I'm not sure from my solution ,but I think it's convincing . Extend AB and let M the intersection of [BA) and the cCFircle (A,AC) MC is parallel to the besictor of BAC (Thales ) then CMA=A/2. And we have DMC =A . so DMA=A+A/2=3A/2 (1) . Now entend [BE) to Z . we have ZED=EDB+EBD =A+A/2 =3A/2 (2). from 1 and 2 we conclude DEBM is cyclic It remains to prove that E,C and M are collinear, but we have CMB =BDE and DBE= CMD ,so they must be collinear . and we are done . Note : if my way is wrong please tell me .

use angle form of ceva click to see more detail

May you tell me the source of this problem? Thank you so much.

toanhocmuonmau123 wrote: May you tell me the source of this problem? Thank you so much. Sorry, I have not read carefully.

EHIC isn't cyclic. Because by angle chasing AC is perpendicular to line BE. Then GCIH is cyclic

Best regards, sunken rock

Nice one, but pretty tricky, because what you need to do is take de perpendicular bisector of $CD$ and let it intersect $ED$ in $G$ and $BE$ in $H$. Also, let $P=EC \cap AB$. Notice that we only need to have $BEDP$ cyclic. Obviously $C$ is the orthocenter of $HAB$, and thus $\angle CHE=A=\angle CGE=\angle GBC + \angle GCB$. Thus $ECGH$ is cyclic and as $\angle CHA=90-2A$ (because $C$ is the orthocenter) this implies $\angle CED=\angle 90-2A$ and as $\angle CGE=90-3A/2$, we have $\angle PCA=A/2$ which implies $EC$ parallel to the $A-$ bisector and thus $BEDP$ cyclic. This sol is not fully mine; I just posted it here because the ones I've seen here were wrong/incomplete

trying_to_solve_br wrote: $\angle CHA=90-2A$ (because $C$ is the orthocenter) this implies $\angle CED=\angle 90-2A$. You can end here. Also there's no need to prove $BEDP$ cyclic.

My solution is a trig bash: Let angle CED=x angle CDE=A/2 so, angle ECB=x+A/2 angle CBE=A So, angle BEC=180°-{(3A/2)+x} AC=AD, C=A+90° B=90°-2A, angle CAD=2A In ∆ABD, sin 3A/ BD= sin(90°-2A)/AD In ∆ABC, sin A/ BC= sin(90°-2A)/AC sin 3A/ BD= sin A/ BC So, if we just evaluate it, CD/ BC=2 cos 2A In∆ECD, EC=CD sin(A/2) / sin x In∆BEC, EC=BC sin A / sin{(3A/2)+x} After evaluation, 2 cos 2A sin{(3A/2)+ x}= 2 sin x cos A/2 sin{(7A/2)+ x} - sin(x+A/2}=0 2cos(x+2A) sin(3A/2)=0 If sin( 3A/2)=0 A≠0 3A/2=180° A=120° C=210°( not possible) cos(x+2A)= 0 x+2A=90° x=90°-2A=B x=angle CED So, angle CED= B @ Krishijivi

Let $K$ be on the extension of $BA$ such that $AC=AD=AK$. (This implies $\angle AKC = \angle ACK = \frac{A}{2}$.) Let the tangent to $(ABC)$ at $B$ intersect $KC$ at $E'$. Then $\angle E'KD = \angle CKD = \frac{\angle CAD}{2} = A = \angle E'BD \implies BE'DK$ cyclic. $\angle E'DC = \angle BDE' = \angle BKE' = \angle AKC = \frac{A}{2} \implies E'=E$. Therefore $\angle CED = \angle KED = \angle KBD = \angle ABC$ as desired.

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