i have no idea why this works and other approaches don't[br][math]KM=LM[/math], show [math]NM=OM[/math][br][br]Let [math]H[/math] be intersection of [math]\left(BMC\right)[/math]. First we have [math]DNBH[/math] and [math]ANCH[/math] cyclic (can be verified by angle chasing [math]\angle NDM[/math] and [math]\angle NAM[/math]. This implies [math]NM\cdot MH=AM\cdot MC=KM\cdot ML[/math], so we have [math]\frac{KM}{NM}=\frac{MH}{ML}[/math][br][br]Next, [math]\frac{OH}{OL}=\frac{OH}{OC}\cdot\frac{OC}{OL}=\frac{OB}{OM}\cdot\frac{OK}{OB}=\frac{OK}{OM}[/math][br]Subtract [math]1[/math] from [math]\frac{OH}{OL}[/math] and [math]\frac{OK}{OM}[/math] to get [math]\frac{LH}{OL}=\frac{KM}{OM}=\frac{LM}{OM}[/math][br][br]If we have [math]\frac{LH}{OL}=\frac{LM}{OM}[/math] then that is also equal to [math]\frac{LH+LM}{OL+OM}=\frac{MH}{ML}[/math][br]But we established [math]\frac{KM}{NM}=\frac{MH}{ML}[/math] which implies [math]\frac{LM}{OM}=\frac{KM}{NM}[/math], which gives us [math]OM=NM[/math].[br][br]Very Weird[br]