Van Khea's Fourth Identity in Triangle

Let $O$ be a point inside $\Delta ABC.$ Lines $AO,BO,CO$ cross the sides of the triangle at $D,E,F,$ as shown:

Van Khea's Fourth Identity in Triangle, problem

Line $MN,$ with $M\in AB$ and $M\in AC$ intersects $AD$ in $P.$ Prove that


$\displaystyle \frac{BM}{MA}\cdot\frac{AE}{EB}+\frac{CN}{NA}\cdot\frac{AF}{FC}=\frac{AO}{OD}\cdot\frac{DP}{PA}.$

(Note that in (*) and also below all segments are thought to be signed.)


The case $MN\parallel BC$ being trivial, we assume without loss of generality that $MN$ meets BC in $Q$ such that $B\in QC.$

Van Khea's Fourth Identity in Triangle, problem

By van Aubel's theorem,



By Menelaus' theorem in $\Delta ABD$ and the transversal $PMQ,$



and in $\Delta ACD$ and the transversal $PNQ,$



Substituting (1-3) into (*) results in an equivalent identity:


which, in turn, can be transformed into




which is equivalent to


The latter is the Ceva condition for $\Delta ABC$ and the cevians through $O.$


The problem has been sent to me by Leo Giugiuc with credits to Van Khea. The proof is by Leo Giugiuc.

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