Let $F$ be intersections of $BB$ and $EC$ , $G$ be intersections of $BF$ and $AC$ , $H$ be intersections of $BF$ and $AE$ then $$\angle HBE = \angle DBE - \angle HBD = \angle DCA - \angle DAB = \angle CAB - \angle DAB = \angle CAD = \angle HBE$$Since $\overline{BE} // \overline {AG}$ Thus $\square ABEG$ is isosceles trapezoid Consider $$\angle AGE = \angle GAB = \angle CAB = \angle ACB$$Implies, $\overline{GE} // \overline{BC}$ Thus $\square CBEG$ is parallelogram Hence, $F$ is midpoint of $EC$ $\blacksquare$

Let $BB$ and $EC$ meet at $M$. Let parallel through $B$ to $EC$ meets $(B D A)$ at point $K$ and $AC$ at point $T$. And parallel through $B$ to $AC$ meets $(B D A)$ at point $N$. Then we have: $\angle BCA=\angle EBC=\angle DKN$ and $\angle DNK=\angle CBT$ which means that $DKN$ and $TBC$ are similar triangles then we have $NK/KD=BC/CT=AB/EB$ (We used that $BECT$ is parallelogram). And from angle chasing we have: $\angle BAC=\angle BCA=\angle EBC=\angle DAN$ which means that $\angle BEA=\angle EAC=\angle BAN=\angle BDN$ $(1)$ and we know that $\angle BAE=\angle BND$ $(2)$. Using $(1)$ and $(2)$ we obtain that $ABE$ and $NBD$ are similar which means that: $NK/KD=AB/EB=NB/BD$, implying that quadriletiral $KNDB$ is harmonic, then $-1=(N,D;B,K)=$we will project it to line $EC$ using vertex $B$ $=(E,C;M,P\infty)=-1$, which completing the solution.

Suppose $BC$ intersects $EC$ at $X$, now $\angle BAD = \angle BAE = \angle CBX = x$ and $\angle CAD = \angle CAE = \angle XBC=y$. By ratio lemma, $\frac{EX}{XC} = \frac{BE}{BC} \cdot \frac{\sin{y}}{\sin{x}}$, so it suffices to prove that $\frac{BE}{BC} = \frac{\sin{x}}{\sin{y}}$. Again by ratio lemma, notice that $\frac{BD}{CD} = \frac{AB}{AC} \cdot \frac{\sin{x}}{\sin{y}}$. Thus it suffices to prove that $\frac{BD}{CD}=\frac{AB}{AC} \cdot \frac{BE}{BC}$ which is the same as proving $\frac{BD}{AB}=\frac{CD}{AC} \cdot \frac{BE}{AB}$ which is just true by sine law on $\bigtriangleup ABE$, $\bigtriangleup ABD$ and $\bigtriangleup ACD$. $\blacksquare$

Let the tangent to the circumcricle at $B$ intersects $EC$ at $F$ Note that $\angle EBF=\angle EBC-\angle FBC=\angle EAC$ Hence $\dfrac{EF}{FC}=\dfrac{BE\cdot \sin \angle EBF}{BC\cdot \sin \angle FBC}=\dfrac{BE\cdot \sin \angle BEA}{BA\cdot \sin \angle BAE}=1$ Which means that $EF=FC$

My Handwritten Solution : The first angle equation is obtained by using tangency criteria too and other properties.

kamatadu wrote: A point $D$ lie on the lateral side $BC$ of an isosceles triangle $ABC$. The ray $AD$ meets the line passing through $B$ and parallel to the base $AC$ at point $E$. Prove that the tangent to the circumcircle of triangle $ABD$ at $B$ bisects $EC$. Well I did solve this that day, was extremely bored You just construct $F$ such that $BEFC$ is a parallelogram, obviously it lies on $AC$ and also we get that $ABEF$ is an isosceles trapezoid so $$\angle FBC=\angle BFE=\angle BAE$$so the tangent($BF$) passes through the midpoint of $EC \blacksquare$

More than half of the 9th graders from India bashed this problem in the contest.I still literally have no idea why Also my solution was the most obvious one as in #2,however cute angle chases seem to have phantom point replacements as in #8,#5 also @starchan orz

HoRI_DA_GRe8 wrote: More than half of the 9th graders from India bashed this problem in the contest.I still literally have no idea why Also my solution was the most obvious one as in #2,however cute angle chases seem to have phantom point replacements as in #8,#5 also @starchan orz probably because people didn't want to take any chances after how terrible day 1 went

Construct a point $G$ on ray $AC$ such that $GBCE$ is a parallelogram. Then, we have $\overline{EB} \parallel \overline{GA},$ and $GE = BC = AB,$ so $GEBA$ is a CyclicISLscelesTrapezoid, which means that $\angle EGB = \angle EAB,$ but $\angle EGB = \angle GBC,$ but $\overline{GB}$ bisects $\overline{EC},$ because diagonals of parallelogram bisect each other, so $\overline{GB}$ is tangent to $(ABD)$ as well as bisects $\overline{CE},$ so we're done. EDIT: bruh this is isomorphic to BVKRB's sol, ig geomains think alike? (although im not a geomain now)