Quote: Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$, and let $K$ be a point on the ray $BA$ beyond $A$. The line $KM$ intersects the side $BC$ at the point $L$. $P$is the point on the segment $BM$ such that $PM$ is the bisector of the angle $LPK$. The line $\ell$ passes through $A$ and is parallel to $BM$. Prove that the projection of the point $M$ onto the line $\ell$ belongs to the line $PK$. Let $H\equiv$ $\ell \cap PK$ and $M'\equiv$ $\ell \cap KL$. The parallel line with $PL$ through $H$ intersects $KL$ at $M'$. Then $\ell$ will be the angle bisector of $\angle L'HK$. Since $\frac {KL'}{KL}$ $=\frac {KH}{KP}$ $=\frac {KA}{KB}$; hence $AL'$ $\|$ $BC$ $\equiv$ $LC$ $\Longrightarrow$ $M$ will be also the midpoint of $LL'$. Or in other words, $ML=ML'$. We have: $\frac {M'L'}{M'K}=\frac {HL'}{HK}=\frac {PL}{PK}=\frac {ML}{MK}=\frac {ML'}{MK}$. As a result, $(K,L',M',M)=-1$, which leads to the fact that $(HK,HL',HM',HM)$ is a harmonic bundle $\Longrightarrow$ $MH\perp \ell$. Our proof is completed then. $\square$

Use Homotety

Attachments: