Quadrilateral

[ins-]

It is given convex quadrilateral ABCD. Let K, L, M, N are the middles of the sides AB, BC, CD, DA respectively. The line KM intersects diagonals AC and BC at the points P and Q. If X is the intersection point of the diagonals and Y is the middle of PQ prove that LN||XY.

[Dimoniada]

Извиняюсь, возможно поторопилась.

[ins-]

It is normal ... problem is a little strange.

[gris]

Ну разве что когда отрезок $XY$ вырождается в точку. Кстати, там небольшая описка: diagonal $BD$.
Упс. О чём это я?

[ins-]

gris you are right the problem is not cleared enough but what should we prove is clear.

[gris]

I should say it's too clear
Хотя по теме "подобие треугольников" как раз.

[ins-]

Why "too" clear? Can it be generalized?
I think the problem can be solved in at least two ways.
Analythical geometry can also be used. It remains me
a problem about Gauss line which I know few solutions.
But the question here is not what technique to use but "how" to use it.

[gris]

Для школьника, постоянного решающего геометрические задачи, подобие $\Delta KLM\,\sim\,\Delta PXQ$ вместе с их медианами бросится в глаза.
Останется оформить и уладить проблемы с вырожденными случаями.
Задача хороша тем, что напоминает о многих теоремах.

[ins-]

Thank you for the solution. The problem is very easy but when you see it independently.
Great solution.
1. Have you ever seen the problem?
2. What is the level of difficulty of this problem?

[Dimoniada]

Из $XQ/QD=XP/CP$ и $XP/PA=XQ/BQ$ сдедует $XP/XQ=AC/BD=KL/LM$, откуда и получается подобие тр-иков. Первые 2 равенства можно получить, вспомнив т.Гаусса для четырёхугольников $ABX(BP \cap AQ)$ и $DCX(CQ \cap DP)$. Задача действительно олимпиадная, для продвинутых школьников

[ins-]

It can be proved the triangles are similar using middle lines, <A+<B+<C=180 and triangles are similar by 3 angles. If we continue to prove that medians are parallel we can use triangles are similar by common angle and corresponding sides equal ratios. My explanation is not good but we don't need to use Gauss line.

Have you ever seen somewhere the problem proposed by me?

[Dimoniada]

Приведите пожалуйста, если не затруднит, ваше решение, ins-. Хочется посмотреть на более короткий путь.

[ins-]

I should go, but tonight I'll write it.

[ins-]

Dear Dimoniada,
I'm sorry for the delay. As I promised I'll explain my idea and probably gris' idea.
I'm using gris' picture.

If we take a look at the triangle ACB - K is the middle of AB, L is the middle of BC.
KL is middle line in ACB => KL ||AC and PX.
By analogy LM||QX.
It follows that <LKM=<XPQ and <LMK=<XQP.
Using that the sum of the angles in the triangle is equal to 180 degrees it follows
that <KLM=<PXQ.
From the equal angles it follows that the triangles KLM and PXQ are similar by three
equal angles.
From the similarity we have:
1) <LKM=<XPQ
2) KL/PX=KM/PQ=(1/2*KM)/(1/2*PQ)=KY/PZ
Where Z is the middle of KM. (the quadrilateral KLMN is parallelogram
and its diagonals have common middle - in our case Z - Z is intersection point of KM and LN)
From 1) and 2) it follows that the triangles KLZ and PXY are similar.
From this similarity it follows that <PYX=<KZL as corresponding angles.
So XY||LN. The statement is proven.

It is probably not the shortest way possible but It keeps the solution elementary
and we don't need to know Gauss line.
I don't have extended math education it is the reason for me to prefere simplest sollutions possible.
If you have any questions, please ask them.

1. Have you ever seen the problem?
2. What is the level of difficulty of this problem?

Are mine questions to all the users.
Thank you for showing me your excellent ideas.

[Dimoniada]

Wow! Что-то со мною не то Всё так элементарно оказалось.