Can someone post a diagram?

PenguinIntegral wrote: Can someone post a diagram? Yes please... And can anybody explain Erdosh Mordelli's Theorem????

Erdos-Mordell Theorem Also here is an image of the problem

can anyone come up with a solution for this? thx

the problem was also posted and solved here.

[asy][asy] import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 11.92, ymin = -2.22, ymax = 6.3; /* image dimensions */ /* draw figures */ draw((-2.9,4.86)--(-2.1,-0.52)); draw((-2.1,-0.52)--(9.04,-0.58)); draw((9.04,-0.58)--(-2.9,4.86)); draw(circle((-0.36,1.49), 2.02)); draw((-3.49,2.18)--(9.04,-0.58)); draw((0.22,2.16)--(-2.1,-0.52)); draw((-2.5,2.17)--(3.07,2.14)); draw((-3.49,2.18)--(3.07,2.14)); /* dots and labels */ dot((-2.9,4.86),dotstyle); label("$A$", (-2.82,4.98), NE * labelscalefactor); dot((-2.1,-0.52),dotstyle); label("$B$", (-2.42,-0.6), SW * labelscalefactor); dot((9.04,-0.58),dotstyle); label("$C$", (9.12,-0.46), NE * labelscalefactor); dot((-0.36,1.49),dotstyle); label("$I$", (-0.28,1.12), NE * labelscalefactor); dot((-2.5,2.17),dotstyle); label("$M$", (-2.86,2.3), NE * labelscalefactor); dot((3.07,2.14),dotstyle); label("$N$", (3.16,2.26), NE * labelscalefactor); dot((-3.49,2.18),dotstyle); label("$L$", (-3.42,2.3), NE * labelscalefactor); dot((0.22,2.16),dotstyle); label("$K$", (0.3,2.28), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] From $MN\parallel CB$ and $IB$ bisecting $\angle ABC$ we have $\angle MKB=\angle MBK$, which implies $MK=BM=\frac{1}{2}AB$. Similarly we have $LN=NC=\frac{1}{2}AC$. Also, since $M$ and $N$ are mid-points, we have $MN=\frac{1}{2} BC$. Now by trianlgle inequality, we have $AI+BI > AB$, $BI+CI>BC$, $CI+AI>CA$, which gives \[AI+BI+CI>\frac{1}{2}(AB+BC+CA) = MK+MN+LN = (LK-LM)+MN+(LM+MN) = 2MN + LK= BC+KL.\] Therefore, we have $AI+BI+CI>BC+KL$. Note: the last step assumes a configuration where $L$ is on the left of $M$ and $K$ is between $M$ and $N$, but the same argument works for other configurations as well.

its easy proble since the diameter is the largest segment in the circle AI>KL and BI+CI>BC we summarize them: AI+BI+CI>BC+KL

try to prove yourself that the quadrilateral KALI-is inscribed

oops this is wrong