# An Identity in Triangle

Let $O$ be a point inside $\Delta ABC.$ Lines $AO,BO,CO$ cross the sides of the triangle at $D,E,F,$ and the lines $EF,DF,DE$ at $P,Q,R,$ respectively.

Prove that

\begin{align}\displaystyle \frac{OP}{AP}\cdot\bigg(\frac{AF}{BF}+\frac{AE}{CE}\bigg)&+\frac{OQ}{BQ}\cdot\bigg(\frac{BF}{AF}+\frac{BD}{CD}\bigg)\\ &+\frac{OR}{CR}\cdot\bigg(\frac{CD}{BD}+\frac{CE}{AE}\bigg)=2. \end{align}

### Proof 1

Apply Menelaus' theorem in $\Delta ABO$ to obtain $\displaystyle\frac{BE}{EO}\cdot\frac{OP}{AP}\cdot\frac{AF}{BF}=1,$ or,

$\displaystyle\frac{OP}{AP}\cdot\frac{AF}{BF}=\frac{EO}{BE}.$

Similarly, in $\Delta ACO,$ $\displaystyle\frac{OP}{AP}\cdot\frac{AE}{CE}=\frac{FO}{CF}.$ It thus follows that

$\displaystyle\frac{OP}{AP}\cdot\bigg(\frac{AF}{BF}+\frac{AE}{CE}\bigg)=\frac{EO}{BE}+\frac{FO}{CF}.$

In the same way.

$\displaystyle\frac{OQ}{BQ}\cdot\bigg(\frac{BF}{AF}+\frac{BD}{CD}\bigg)=\frac{DO}{AD}+\frac{FO}{CF}$ and

$\displaystyle\frac{OR}{CR}\cdot\bigg(\frac{CD}{BD}+\frac{CE}{AE}\bigg)=\frac{EO}{BE}+\frac{DO}{AD}.$

Summing up gives

\begin{align}\displaystyle \frac{OP}{AP}\cdot\bigg(\frac{AF}{BF}+\frac{AE}{CE}\bigg)&+\frac{OQ}{BQ}\cdot\bigg(\frac{BF}{AF}+\frac{BD}{CD}\bigg)\\ &+\frac{OR}{CR}\cdot\bigg(\frac{CD}{BD}+\frac{CE}{AE}\bigg)\\ &=2\bigg(\frac{DO}{AD}+\frac{EO}{BE}+\frac{FO}{CF}\bigg)\\ &=2, \end{align}

because, by Gergonne's theorem, $\displaystyle\bigg(\frac{DO}{AD}+\frac{EO}{BE}+\frac{FO}{CF}\bigg)=1.$

### Proof 2

With $[X]$ denoting the area of figure $X,$ let $[\Delta BCO]=x,[\Delta ACO]=y, [\Delta ABO]=z.$ In $\Delta AEF,$ the cevians $AP,BE,CF$ are concurrent at $O.$ So, by van Aubel's theorem, $\displaystyle\frac{AO}{OP}=\frac{AB}{BF}+\frac{AC}{CE}.$ Also, $\displaystyle\frac{AF}{BF}=\frac{y}{x}$ because the altitudes from $A$ and $B$ to the common base $C$ are proportional to $AF$ and $BF.$ Therefore, $\displaystyle\frac{AB}{BF}=\frac{x+y}{x}.$ Similarly, $\displaystyle\frac{AC}{CE}=\frac{x+z}{x}.$ It follows that $\displaystyle\frac{AO}{OP}=\frac{2x+y+z}{x},$ and $\displaystyle\frac{AP}{OP}=\frac{x+y+z}{x}$ and $\displaystyle\frac{OP}{AP}=\frac{x}{x+y+z}.$

On the other hand, $\displaystyle\frac{AF}{BF}=\frac{y}{x}$ and $\displaystyle\frac{AE}{CE}=\frac{z}{x},$ implying $\displaystyle\frac{AF}{BF}+\frac{AE}{CE}=\frac{y+z}{x}$ and, subsequently, $\displaystyle\frac{OP}{AP}\bigg(\frac{AF}{BF}+\frac{AE}{CE}\bigg)=\frac{y+z}{x+y+z}.$

Similarly, $\displaystyle\frac{OQ}{CQ}\bigg(\frac{CE}{AE}+\frac{CD}{BD}\bigg)=\frac{x+z}{x+y+z}$ and $\displaystyle\frac{OR}{BR}\bigg(\frac{BD}{CD}+\frac{BF}{AF}\bigg)=\frac{x+y}{x+y+z}.$ All that remains is to add up the three latter identities.

### Acknowledgment

The problem has been posted by Leo Giugiuc at the CutTheKnotMath facebook page with participation of Van Khea. I figure that the first proof is more Van Khea's, while the second is more Leo's.