Points $K$ and $L$ are taken on extension of the bases $AD\; and \; BC$ of trapezoid $ABCD$ beyond $A\; and \; C$, respectively. Line segment $KL$ intersects sides $AB$ and $CD$ at $M\; and\; N$ respectively, also intersects diagonals $AC\; and\; BD$ at $O \; and\; P$ respectively. Prove that if $KM = NL$, then $KO = PL$.
I am able to draw the picture. Here it is -
This is a problem from "Problems in plane and solid geometry" book. There is a solution in the book but I didn't understand the solution. So I posted it here. I am just getting things messed up. I badly need a full solution.
My Work
I can only find that -
$PL:PK = BL:KD$ and $OK:OL = KA:CL$.
Now how to proceed.
You have $PL:PK = BL:KD$ and $OK:OL = KA:CL$
or $\frac{LP}{LB} = \frac{KP}{KD}$ and $\frac{KO}{KA} = \frac{LO}{LC}$, giving $\frac{LP}{KO} = \frac{LB}{KA}\frac{KP}{KD}\frac{LC}{LO}$
Also $\frac{LN}{LC} = \frac{KN}{KD}$ and $\frac{KM}{KA} = \frac{LM}{LB}$
Now given $KM=LN$ (which also gives $KN=LM$)
Thus $LC\frac{KN}{KD} = KA \frac{LM}{LB} \implies \frac{LC}{KD}\frac{LB}{KA} = \frac{LM}{KN} = 1$ so $\frac{LP}{KO} = \frac{KP}{LO} = \frac{KL - LP}{KL - KO}$
$\begin{align}\text{Then}\quad LP(KL-KO) &= KO(KL-LP) \\ LP\cdot KL &= KO\cdot KL \\ \text{and}\quad LP&=KO \quad\text{ as required.}\end{align}$