Proof for an equality in a triangle

Question

$O$ is the point inside triangle $ABC$ . The lines joining the three vertices $A,B,C$ to $O$ cut the opposite sides in $K,L$, and $M$ respectively. A line through $M$ parallel to $KL$ cuts the line $BC$ at $V$ and $AK$ at $W$. Prove that $VM=MW$. I tried using the ceva's theorem in triangle $ABC$ and equating it with the relation obtained by Menellau's theorem . Then I used basic proportionality theorem and after using them I just got one relation. $$\frac {BK}{BV}=\frac {AL}{AP}$$

Now I am stuck please help me.

Answer 1

Construct a parallel line to $KL$ passing through $A$. Let it intersect $BC$ at $E$. Then from the parallel lines we have:

$$\frac{KE}{VE} = \frac{KA}{AW}$$

Now from Ceva's Theorem on the $\triangle ABC$ and $O$ we have:

$$1 = \frac{AM}{MB} \times \frac{BK}{KC} \times \frac{CL}{LA} = \frac{AM}{MB} \times \frac{BK}{KC} \times \frac{CK}{KE} \implies \frac{AM}{MB} = \frac{KE}{BK}$$

Now from Menelaus Theorem on $\triangle ABK$ and line $V-M-W$ we have:

$$1 = \frac{AM}{MB} \times \frac{BV}{VK} \times \frac{KW}{WA} = \frac{KE}{KB} \times \frac{BV}{VK} \times \frac{KW}{WA} \implies \frac{BK}{BV} = \frac{KE}{VK} \times \frac{KW}{AW} = \frac{KA}{KW} \times \frac{KW}{AW}$$

$$ \implies \frac{BK}{BV} = \frac{KA}{KW}$$

Finally from the Menalaus' Theorem on $\triangle VWK$ and line $A-M-B$ we have:

$$1=\frac{VB}{BK} \times \frac{KA}{AW} \times \frac{WM}{MV} \implies \frac{WM}{MV} = 1 \implies MV = MW$$

Therefoere we have that $M$ is the midppint of $WV$. Hence the proof.

Answer 2

Let $T$ be the intersection of $KL$ and $OC$. Since $KL$ is parallel to $VW$, we have $$VM/KT=CM/TC,\quad MW/KT=OM/OT.$$Hence it suffices to show that $$CM/CT=MO/OT\iff CT/OT=CM/MO.$$On the other hand, note that the line KTL intersects with the triangle COB, by Menellau's theorem, we have $$ CT/OT \cdot OL/LB \cdot BK/KC=1. $$ Hence $$\frac{CT}{OT}=\frac{LB\cdot KC}{OL\cdot BK}.$$ Let $S_{ABC}$ be the area of the triangle $ABC$ and so on. Then $$ \frac{LB}{OL}=\frac{S_{ABC}}{S_{AOC}},\frac{CM}{MO}=\frac{S_{ABC}}{S_{AOB}},\frac{KC}{BK}=\frac{S_{AOC}}{S_{AOB}} $$ Therefore $$ \frac{CM}{MO}=\frac{S_{ABC}}{S_{AOB}}=\frac{LB\cdot KC}{OL\cdot BK}=\frac{CT}{OT}. $$We are done.

Answer 3

If you know some projective geometry, then you can easly solve it with cross-ratio. Say line $LK$ cuts line $AB$ at $X$.

\begin{eqnarray*} (V,W;M,\infty) &=& (KV,KW;KM,K\infty)\\ &=& (KA,KB;KM,KX)\\ &=& (A,B;M,X)\\ &=& -1 \end{eqnarray*} The last equality is because quadrilateral $CLOK$ is harmonical. So $M$ must be a midpoint of segment $VW$.

Answer 4

Let $$A=\left(a_x,a_y\right),B=\left(b_x,b_y\right),C=\left(c_x,c_y\right)\\O=\left(o_x,o_y\right),K=\left(k_x,k_y\right),L=\left(l_x,l_y\right)\\M=\left(m_x,m_y\right),V=\left(v_x,v_y\right),W=\left(w_x,w_y\right)$$ where $(x,y)$ represents coordinates of a given point. Now, apply the formula for finding the analytic form of a line $XY$ through two given points $X=(x_1,y_1)$ and $Y=(x_2,y_2)$ $$XY:y-y_1=k_{xy}(x-x_1)\\k_{xy}=\frac{y_2-y_1}{x_2-x_1}$$ where $k$ is the direction coefficient of the line (tangent of their slope angle). Applying this formula to $\triangle_{ABC}$ we get $$AB:y-a_y=\frac{b_y-a_y}{b_x-a_x}\left(x-a_x\right)\\BC:y-b_y=\frac{c_y-b_y}{c_x-b_x}\left(x-b_x\right)\\CA:y-c_y=\frac{a_y-c_y}{a_x-c_x}\left(x-c_x\right)\\AO:y-a_y=\frac{o_y-a_y}{o_x-a_x}\left(x-a_x\right)\\BO:y-b_y=\frac{o_y-b_y}{o_x-b_x}\left(x-b_x\right)\\CO:y-c_y=\frac{o_y-c_y}{o_x-c_x}\left(x-c_x\right)$$ Now, because $K=AO\bigcap BC,L=BO\bigcap CA,M=CO\bigcap AB$ we can calculate the coordinates of $K,L,M$ $$k_y-a_y=\frac{o_y-a_y}{o_x-a_x}\left(k_x-a_x\right)\\k_y-b_y=\frac{c_y-b_y}{c_x-b_x}\left(k_x-b_x\right)$$From this we can calculate $k_x$ and $k_y$ $$k_x=\frac{a_x \left(-b_x c_y+b_x o_y+b_y c_x-c_x o_y\right)+o_x \left(a_y \left(c_x-b_x\right)+b_x c_y-b_y c_x\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x-c_x o_y+c_y o_x}\\k_y=\frac{o_y \left(a_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+a_y \left(c_y \left(o_x-b_x\right)+b_y \left(c_x-o_x\right)\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x-c_x o_y+c_y o_x}$$ We can similarly calculate it for $L$ and $M$ $$l_x=\frac{b_x \left(-c_x a_y+c_x o_y+c_y a_x-a_x o_y\right)+o_x \left(b_y \left(a_x-c_x\right)+c_x a_y-c_y a_x\right)}{b_x \left(c_y-a_y\right)+b_y \left(a_x-c_x\right)+c_x o_y-c_y o_x-a_x o_y+a_y o_x}\\l_y=\frac{o_y \left(b_x \left(c_y-a_y\right)+c_x a_y-c_y a_x\right)+b_y \left(a_y \left(o_x-c_x\right)+c_y \left(a_x-o_x\right)\right)}{b_x \left(c_y-a_y\right)+b_y \left(a_x-c_x\right)+c_x o_y-c_y o_x-a_x o_y+a_y o_x}\\m_x=\frac{c_x \left(-a_x b_y+a_x o_y+a_y b_x-b_x o_y\right)+o_x \left(c_y \left(b_x-a_x\right)+a_x b_y-a_y b_x\right)}{c_x \left(a_y-b_y\right)+c_y \left(b_x-a_x\right)+a_x o_y-a_y o_x-b_x o_y+b_y o_x}\\m_y=\frac{o_y \left(c_x \left(a_y-b_y\right)+a_x b_y-a_y b_x\right)+c_y \left(b_y \left(o_x-a_x\right)+a_y \left(b_x-o_x\right)\right)}{c_x \left(a_y-b_y\right)+c_y \left(b_x-a_x\right)+a_x o_y-a_y o_x-b_x o_y+b_y o_x}$$ Let's continue. Line $MW$ is the line parallel to $KL$. Parallel means that it has the same slope coefficient which means that $$\frac{m_y-w_y}{m_x-w_x}=\frac{k_y-l_y}{k_x-l_x}$$ Therefore the line $MW$ is defined as $$y-v_y=\frac{k_y-l_y}{k_x-l_x}(x-v_x)$$ There is now enough information to calculate the coordinates of $V$ and $W$ $$v_x=\frac{m_x \left(-b_x c_y+b_x w_y+b_y c_x-c_x w_y\right)+w_x \left(m_y \left(c_x-b_x\right)+b_x c_y-b_y c_x\right)}{m_x \left(b_y-c_y\right)+m_y \left(c_x-b_x\right)+b_x w_y-b_y w_x-c_x w_y+c_y w_x}\\v_y=\frac{w_y \left(m_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+m_y \left(c_y \left(w_x-b_x\right)+b_y \left(c_x-w_x\right)\right)}{m_x \left(b_y-c_y\right)+m_y \left(c_x-b_x\right)+b_x w_y-b_y w_x-c_x w_y+c_y w_x}\\w_x=\frac{m_x \left(-b_x c_y+b_x v_y+b_y c_x-c_x v_y\right)+v_x \left(m_y \left(c_x-b_x\right)+b_x c_y-b_y c_x\right)}{m_x \left(b_y-c_y\right)+m_y \left(c_x-b_x\right)+b_x v_y-b_y v_x-c_x v_y+c_y v_x}\\w_y=\frac{v_y \left(m_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+m_y \left(c_y \left(v_x-b_x\right)+b_y \left(c_x-v_x\right)\right)}{m_x \left(b_y-c_y\right)+m_y \left(c_x-b_x\right)+b_x v_y-b_y v_x-c_x v_y+c_y v_x}$$ Now using the fact that $k_{mw}=k_{kl}$ we have

$$\large w_y=\frac{a_x (m_y o_y (c_x-b_x)+c_y (b_x m_y-l_x m_y+l_y m_x-l_y w_x-m_x o_y+o_y w_x)+b_y (-c_x m_y+l_x m_y+l_y (w_x-m_x)+m_x o_y-o_y w_x)+a_y (b_x (-c_y m_x+c_y w_x-l_x m_y+l_y m_x-l_y w_x+m_y o_x)+b_y (c_x-o_x) (m_x-w_x)+c_x l_x m_y-c_x l_y m_x+c_x l_y w_x-c_x m_y o_x+c_y m_x o_x-c_y o_x w_x)+b_x c_y m_x o_y-b_x c_y m_y o_x-b_x c_y o_y w_x+b_x l_x m_y o_y-b_x l_y m_x o_y+b_x l_y o_y w_x-b_y c_x m_x o_y+b_y c_x m_y o_x+b_y c_x o_y w_x-b_y l_x m_y o_x+b_y l_y m_x o_x-b_y l_y o_x w_x-c_x l_x m_y o_y+c_x l_y m_x o_y-c_x l_y o_y w_x+c_y l_x m_y o_x-c_y l_y m_x o_x+c_y l_y o_x w_x}{a_x (b_x c_y-b_x o_y+b_y (l_x-c_x)+c_x o_y-c_y l_x)-a_y (b_x-c_x) (l_x-o_x)-b_x c_y o_x+b_x l_x o_y+b_y c_x o_x-b_y l_x o_x-c_x l_x o_y+c_y l_x o_x}$$

And we can do similarly for $w_x$ and also for point $V$. Now we can substitute $k_x$ and $k_y$ back:

$$\Large \large v_x=\frac{-\left(\frac{a_x \left(-b_x c_y+b_x o_y+b_y c_x-c_x o_y\right)+o_x \left(a_y \left(c_x-b_x\right)+b_x c_y-b_y c_x\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x-c_x o_y+c_y o_x}-l_x\right) \left(m_y-v_y\right)+\frac{o_y \left(a_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+a_y \left(c_y \left(o_x-b_x\right)+b_y \left(c_x-o_x\right)\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x+\frac{o_y \left(a_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+a_y \left(c_y \left(o_x-b_x\right)+b_y \left(c_x-o_x\right)\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x-c_x o_y+c_y o_x}\left(\frac{m_y-w_y}{m_x-w_x}-\frac{k_y-l_y}{k_x-l_x}\right)-c_x o_y+c_y o_x} m_x-l_y m_x}{\frac{o_y \left(a_x \left(b_y-c_y\right)+b_x c_y-b_y c_x\right)+a_y \left(c_y \left(o_x-b_x\right)+b_y \left(c_x-o_x\right)\right)}{a_x \left(b_y-c_y\right)+a_y \left(c_x-b_x\right)+b_x o_y-b_y o_x-c_x o_y+c_y o_x}-l_y}$$ Do the same for $v_y,w_x,w_y$ (I'm not posting these just to save space). After finding coordinates of $V$ and $W$ we can very easily show that $$\sqrt{(v_x-m_x)^2+(v_y-m_y)^2}=\sqrt{(w_x-m_x)^2+(w_y-m_y)^2}$$ which is exactly what we need (it means that length of $VM$ is the same as the length of $WM$).