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

(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 \(AQ = D_2P\).

(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.

(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 \(MX \perp BC \).

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## Comments

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TopNewestQ2. 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 \(AD\).Log in to reply

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yes i want soutions please post the solutions.

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