# Synthetic Geometry Group - Alan Yan's Proposal

Note that although these questions are marked as "Synthetic", you are allowed to solve them with any method you want.

1. (Medium Difficulty) Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides BC and AC, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex A is denoted by Q. Prove that $AQ = D_2P$.

2. (Easy Difficulty) Let triangle $ABC$ satisfy $2BC = AB + AC$ and have incenter I and circumcircle $\omega$. Let D be the intersection of $AI$ and $\omega$ (with A, D distinct). Prove that I is the midpoint of AD.

3. (Hard Difficulty) Let the incircle $\omega$ of triangle $ABC$ touch $BC, CA,$ and $AB$ at $D , E$ and $F$, respectively. Let $Y_1 , Y_2, Z_1, Z_2$ and $M$ be the midpoints of $BF , BD, CE, CD$ and $BC$, respectively. Let the intersection of $Y_1Y_2$ and $Z_1Z_2$ be $X$. Prove that $MX \perp BC$.

If you want solutions posted, just comment down below.

Note by Alan Yan
3 years, 11 months ago

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Q2. It is clear that $ABDC$ is cyclic. $\angle DBC = \angle DAC = \angle DAB = \angle DCB$. So, $\Delta DCB$ is isosceles and $DC = DB$. But, $\angle DIB = \angle DBI$(by some angle chasing). So, $DI = DB =DC$. But, since $ABDC$ is cyclic, using ptolemy's theorem we get $AD = 2DB = 2 DI$, which implies that $AI = ID$. So, $I$ is mid-point of $AD$.

- 3 years, 11 months ago

Does anyone want solutions?

- 3 years, 10 months ago

yes i want soutions please post the solutions.

- 3 years, 7 months ago