Found problems: 4
In a convex quadrilateral $ABCD$, $\angle ABD=30^\circ$, $\angle BCA=75^\circ$, $\angle ACD=25^\circ$ and $CD=CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE=BD$.
[i]Proposed by BJ Venkatachala[/i]
In a convex quadrilateral $ABCD$ sides $AB$ and $DC$ are both divided into $m$ equal parts by points $A, S_1 , S_2 , \ldots , S_{m-1} ,B$ and $D,T_1, T_2, \ldots , T_{m-1},C,$ respectively (in this order).
Similarly, sides $BC$ and $AD$ are divided into $n$ equal parts by points $B,U_1,U_2, \ldots, U_{n-1},C$ and $A,V_1,V_2, \ldots,V_{n-1}, D$. Prove that for $1 \leq i \leq m-1$ each of the segments $S_i T_i$ is divided by the segments $U_j V_j$ ($1\leq j \leq n-1$) into $n$ equal parts
Let $ABCD$ be a convex quadrilateral with $\angle A+\angle D=90^\circ$ and $E$ the point of intersection of its diagonals. The line $\ell$ cuts the segments $AB$, $CD$, $AE$ and $ED$ in points $X,Y,Z,T$, respectively. Suppose that $AZ=CE$ and $BE=DT$. Prove that the length of the segment $XY$ is not larger than the diameter of the the circumcircle of $ETZ$.
[i]Proposed by A. Kuznetsov, I. Frolov[/i]
Let $\triangle ABC$ have $AB = 1$ cm, $BC = 2$ cm and $AC = \sqrt{3}$ cm. Points $D$, $E$ and $F$ lie on segments $AB$, $AC$ and $BC$ respectively are such that $AE = BD$ and $BF = AD$. The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ADE$ for the second time at $M$ and the angle bisector of $\angle ABC$ intersects the circumcircle of $\triangle BDF$ at $N$. Determine the length of $MN$.