The quadrilateral OLMP (which appears to be a trapezoid) shares the side OP with the triangle KOP.
Area of OLMP=½(OP+LM)XY where XY is the distance between the parallel sides OP and LM and is the height of the trapezoid.
Area of KOP=½OP.KY where KY is the height of the triangle.
These areas are the same so:
½(OP+LM)XY=½OP.KY, and KY=(OP+LM)XY/OP, which can be written KY=(1+LM/OP)XY.
The areas are unaffected by the position of Y on OP. The point U has not been defined in the question, and I think it should be Y, so KYX is the perpendicular, points K (KOP vertex), X and Y lie on the perpendicular line extending from K to X. KY and XY still maintain their roles as heights of KOP and OLMP respectively. Since the positioning of KYX is arbitrary, the lengths OY and LX are also arbitrary, because all that is required is to have OP and LM parallel to one another and XY is the distance between them.
By Pythagoras, KO2=KY2+OY2 and KL2=KX2+LX2.
The next thing to do is to choose a specific example in order to test the validity of the proposed equation. Let's make the following length assignments:
OP=10, LM=5, OY=6, LX=2, XY=8.
Hence KY=(1+5/10)8=12. Area of KOP=½(10×12)=60. Area of OLMP=½(10+5)8=60. The areas are equal.
KO2=144+36=180; KL2=(12+8)2+4=404.
KO2/KL2=180/404=45/101, so KO/KL=0.6675 approx. 1/√2=0.7071 approx. Clearly KO/KL≠1/√2. So the proposed equation is not true, based on the assumption that KUX should be KYX.
Now, suppose OP is shifted horizontally so that OY=8 instead of 6, and no other changes are made, including the areas. KL is still the same length, but KO2=144+64=208 so (KO/KL)2=208/404=0.5148 and KO/KL=0.7175 approximately.
It can be seen that the ratio KO/KL varies.