i have no idea why this works[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 NDM and NAM. This implies [math]NM\cdot MH=AM\cdot MC=KM\cdot ML[/math], so we have [math]\frac{MK}{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 1, we get [math]\frac{LH}{OL}=\frac{KM}{OM}=\frac{LM}{OM}[/math][br][br]if we have [math]\frac{LM}{OM}=\frac{LH}{OL}[/math] then that is also equal to [math]\frac{LM+LH}{OM+OL}=\frac{MH}{ML}[/math][br]but we established that was equal to [math]\frac{ML}{NM}[/math] which implies [math]NM=OM[/math][br]freakin weird[br]is some sort of diagram dependency necessary here