euclidean-geometrygeometryplane-geometry
In $\triangle ABC$ the cevians $AD$,$BE$ and $CF$ are concurrent at the point $O$. $DL\parallel EF$ and $BE$ intersects $DL$ at $K$. How to prove that $DK = DL$.
I applied menelaus theorem twice in the $\triangle KEL$ and got a relation $\frac{DL^2} {DK^2}=\frac{BE} {BK}\cdot \frac{CL} {CE} \cdot \frac{AL} {AE} \cdot \frac{EO} {KO}$ and also tried to use the ratio of similar triangles but no result. any ideas on utilizing the ratio of similar triangles.
Initial condition: In $\triangle ABC$, cevians AA’, BB’, and CC’ meet at P. D is a point on AP.
Through D, draw DF // AC cutting CC’ at F. Similarly draw DE // AB cutting BB’ at E. Join EF.
The first question is “will $\triangle DEF \sim\triangle ABC$?”
Note that, after the construction, we have $\triangle PAC \sim \triangle PDF$ and this yields $\dfrac {PF}{PC} = \dfrac {PD}{PA}$. Similarly, we have $\dfrac {PE}{PB} = \dfrac {PD}{PA}$.
Then, $\dfrac {PF}{PC} = \dfrac {PE}{PB}$. This means $\triangle PEF \sim \triangle PBC$ because $\angle BPC$ is the common angle. Therefore, $\angle PEF= \angle PBC$ and $\angle PFE=\angle PCB$. By AAA, $\triangle ABC \sim \triangle DEF$. In fact, the $\triangle DEF$ so constructed is the only triangle meeting the given condition.
Finally, EF // BC, a result already obtained in the last paragraph.
Further explanation:
For a particular point D on AP. The lines DE an DF so constructed fixed $\angle EDF$ to match $\angle BAC$.
To simply the picture drawing, we let $\angle BAC = 90^0$. For similarity to occur, we need $\angle EDF = 90^0$ also, so that D is against A.
$\triangle DEF$ is then constructed according to way suggest above.
Suppose there is another triangle $DE’F’ $ constructed with (1) E’ on BP; (2)F’ on PC and $\angle E’DF’ = \angle EDF = \angle BAC = 90^0$.
BY considering $\triangle DFX$ and $\triangle DF’X$, we can see that $\angle DF’X \ne \angle DFX = \angle ACB$. This means $\triangle DE’F’$ so constructed will never be similar to $\triangle ABC$.
This is a Projective solution. Refer this if you are new to projective. Also the point $H$ in your Diagram, wasn't necessary, so I removed it and introduce a new point called $H$.
Solution: Define $H = DA\cap BC$.
By $lemma$ $2$ from the handout I linked, we get that it is enough to show that $(H,G;C,B)=-1$.
By $lemma$ $5$ from the handout I linked, we get that if we show that $\angle AGH=90$ and $AG$ bisects $\angle BAC$, then we are done. But as $AD, AG$ are angle bisectors we have $\angle HAG =90$ and by question it's given that $AG$ bisects $\angle BAC$. And we are done.
Best Answer
Following timon92, we are going to prove $\dfrac{DA}{AJ}=\dfrac{DO}{OJ}$.
Consider the figure (only add point $M$, which is an intersection of lines $EF$ and $BC$), we focus on $\triangle MDJ$, and apply Menelaus's theorem 4 times: \begin{align} \color{blue}{\frac{MB}{BD}}\color{red}{\frac{DA}{AJ}}\frac{JF}{FM}&=1;\tag{1}\\ \color{blue}{\frac{MB}{BD}}\color{orange}{\frac{DO}{OJ}}\color{green}{\frac{JE}{EM}}&=1;\tag{2}\\ \color{purple}{\frac{MC}{CD}}\color{orange}{\frac{DO}{OJ}}\frac{JF}{FM}&=1;\tag{3}\\ \color{purple}{\frac{MC}{CD}}\color{red}{\frac{DA}{AJ}}\color{green}{\frac{JE}{EM}}&=1.\tag{4} \end{align} Combining (1) and (4), and combining (2) and (3), respectively, we obtain \begin{align} \left(\color{red}{\frac{DA}{AJ}}\right)^2 =\left(\color{blue}{\frac{MB}{BD}}\frac{JF}{FM}\color{purple}{\frac{MC}{CD}}\color{green}{\frac{JE}{EM}}\right)^{-1} =\left(\color{orange}{\frac{DO}{OJ}}\right)^2. \end{align}