At first, we will find the lengths of LK, Lm, ON, OP, then use them to find the ratios between them
The rule of the distance is
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]For LK
Since L = (3, 6), K = (1,5.33), then
[tex]\begin{gathered} LK=\sqrt{(3-1)^2+(6-5.33)^2} \\ LK=\sqrt{4+0.4489} \\ LK=\sqrt{4.4489} \end{gathered}[/tex]For LM
Since L = (3, 6), M = (5, 6.67), then
[tex]\begin{gathered} LM=\sqrt{(3-5)^2+(6-6.67)^2} \\ LM=\sqrt{4+0.4489} \\ LM=\sqrt{4.4489} \end{gathered}[/tex]For ON
Since O = (3, 2.59) and N = (5, 4.2), then
[tex]\begin{gathered} ON=\sqrt{(3-5)^2+(2.59-4.2)} \\ ON=\sqrt{4+2.5921} \\ ON=\sqrt{6.5921} \end{gathered}[/tex]For OP
Since O = (3, 2.59), P = (1, 0.99), then
[tex]\begin{gathered} OP=\sqrt{(3-1)^2+(2.59-0.99)^2} \\ OP=\sqrt{4+2.56} \\ OP=\sqrt{6.56} \end{gathered}[/tex]Now let us find the ratios between them
[tex]\begin{gathered} \frac{KL}{LM}=\frac{\sqrt{4.4489}}{\sqrt{4.4489}}=1 \\ \frac{PO}{ON}=\frac{\sqrt{6.56}}{\sqrt{6.5921}}=0.9975\approx1 \\ \frac{KL}{LM}=\frac{PO}{ON}=1 \end{gathered}[/tex]That means, Parallel lines intercept equal parts
By joining MP
We will have Triangle KPM
Since KL = LM ------- Proved using the distance formula
Since LQ // KP ------ Given
Then MQ = QP ------- Using the theorem down
The theorem
If a line is drawn from a midpoint of one side of a triangle parallel to the opposite side, then it will intersect the 3rd side in its midpoint (Q is the midpoint of MP)
Parallel lines intercept equal parts