# In a quadrilateral ABCD ,which is not a parallelogramm. On the rays AB,CB,CD,AD we put… [closed]

In a quadrilateral $$ABCD$$, which is not a parallelogram, on rays $$AB$$, $$CB$$, $$CD$$, $$AD$$ we put points $$K$$, $$L$$, $$M$$, $$N$$ such that $$KL\parallel MN\parallel AC$$ and $$LM\parallel KN\parallel BD$$. Prove that line which contains the midpoints of $$AC$$ and $$BD$$ goes through the point of intersection of the diagonals of parallelogram $$KLMN$$.

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• What have you tried? What kind of tools can you use? – Aretino Mar 12 at 18:58
• I tried to find the parallelogram of the Option, calculate the angles, but the problem was never solved. – Yaroslav Mar 12 at 20:03 E is the intersection of KL and BD. I is the midpoint of KL. F is the intersection of LM and AC. J is the midpoint of LM.

As I is midpoint of KL, P is midpoint of AC, similarly as J is the midpoint of LM, Q is the midpoint of BD (important).

$$\dfrac {BE}{BO} = \dfrac{KE}{AO} = \dfrac{NG}{AO} = \dfrac{DG}{DO} = \dfrac{BE+DG}{BO+DO}=\dfrac{BD-EG}{BD} = 1- \dfrac{EG}{BD} =$$

$$1- \dfrac{LM}{BD} = 1- \dfrac{KN}{BD} \text{ , now }$$

$$1- \dfrac{LM}{BD} = 1- \dfrac{CF}{CO} \text{ and } 1- \dfrac{KN}{BD} = 1- \dfrac{AH}{AO} \text{ so }$$

$$1- \dfrac{CF}{CO} = 1- \dfrac{AH}{AO} =$$

$$\dfrac{CO-CF}{CO} = \dfrac{AO-AH}{AO} =$$

$$\dfrac{OF}{CO} = \dfrac{HO}{AO} = \dfrac{OF+HO}{CO+AO} = \dfrac{HF}{AC} = \dfrac{KL}{AC}$$

To summarize some of earlier steps.

$$1- \dfrac{LM}{BD} = 1- \dfrac{CF}{CO} = \dfrac{OF}{CO} = \dfrac{HF}{AC} = \dfrac{KL}{AC}$$

So,

$$\dfrac{LM}{BD} + \dfrac{KL}{AC}= 1$$

So,

$$\dfrac{LM/2}{BD/2} + \dfrac{KL/2}{AC/2}= 1$$

$$\dfrac{LJ}{BQ} + \dfrac{LI}{CP}= 1$$

Now from "I" draw a parallel line to BD so parallel to LM, Suppose this line intersects PQ at X and from J draw a parallel line to AC so parallel to KL, Suppose this line intersects PQ at Y and let us assume these two points X and Y are different. Now

$$\dfrac{LJ}{BQ} = \dfrac{CJ}{CQ} = \dfrac{PY}{PQ} \text{ and }\dfrac{LI}{CP} = \dfrac{BI}{BP} = \dfrac{PX}{PQ} \text{ so }$$

$$\dfrac{PY}{PQ} + \dfrac{PX}{PQ} = 1 \text{ so } = PY + PX = PQ$$

This is only possible when X and Y both are same. And we are done as considering this point X, we can see X is the intersection of the diagonals of KLMN (IXJL is a parallelogram and I is the midpoint of KL).

Note: I drew the picture in power point so not very accurate. Any suggestion welcome.

deleted Apr 2 at 1:52

Let $$r = AK/AB$$. Then, by similar triangles, $$CL/CB = CM/CD = AN/AD = r$$ as well.

This illustration shows the situation with $$0, but neither the geometry nor the algebra requires such a restriction. Treating all the points as $$2$$-dimensional vectors, we have the following equations: \begin{align} K &= A + r(B - A)\\ L &= C + r(B - C)\\ M &= C + r(D - C)\\ N &= A + r(D - A) \end{align}

Adding these equations, $$K + L + M + N = (2-2r)(A + C) + 2r(B + D),$$ or $$\frac14 (K + L + M + N) = (1-r)\left(\frac12(A + C)\right) + r\left(\frac12(B + D)\right).$$ The left-hand side is the center of parallelogram KLMN; the right-hand side is a point on the line joining the midpoint of AC and the midpoint of BD.

deleted Apr 2 at 1:52