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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 -
enter image description here
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.

1 Answers 1

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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 $\frac{LP}{KO} = \frac{KP}{LO} = \frac{KL - LP}{KL - KO}$

$\begin{align}\text{Then}\quad LP(KL-KO) &= KO(KL-LP) \\ LP\cdot KL &= KO\cdot KL \\ \text{and}\quad LP&=KO \quad\text{ as required.}\end{align}$

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    I might also look for a more geometric argument. – 2017-01-21
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    This seems more harder than the book's solution :| – 2017-01-21
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    Very likely; I was just starting from the ratios you gave and pushing forward with calculations. – 2017-01-21
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    This solution doesn't uses geometry at all ! :O :O But somehow this solves this problem ! But this is nice solution (y) – 2017-01-21
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    It uses geometry to get the ratio equalities from similar triangles, and a little bit to get line ratios, but yes, it's not very geometric. – 2017-01-21