Parallelogram

[Amicci]

Hello there lovely mathematicians, this is my first time here

I was helping my younger brother with an exercise, but i just couldn't solve it, and i would appreciate some guidance.
Let ABCD be a parallelogram, the points I and J be respectively the midpoints of [DC] and [BC], the line (IJ) intersects with (AD) in E. Prove that DE = CJ .
PS: i couldn't use any of the theorems i knew, such as the BPT ( Basic Proportionality Theorem, also known as Thales Theorem ), because my brother hasn't studied it yet, so i could use only basic beginner middle school geometry, such as the properties of parallelograms, and central/axial symmetry.

 

[mmm4444bot]

… i just couldn't solve it
… i couldn't use any of the theorems i knew
… i could use only basic beginner middle school geometry …

Welcome to the boards, Amicci. It will be helpful for you to post your attempt, so that tutors do not need to start from scratch. (The issue might merely be a relatively-minor misstep.) Thanks!

?

 

[HallsofIvy]

Can us ugly mathematicians help? You should see that segments EI and IC are congruent (since I is a midpoint), that angles DIE and ICJ are congruent ("vertical angles"), and that angles EDI and ICJ are congruent ("alternate interior angles") so that triangles EDI and OCJ are congruent (ASA).

 

[Amicci]

Okay, here goes my attempts:
1) In the triangle BCD, the line (IJ) passes through the midpoints of two of his edges [DC] and [BC], thus the line (IJ) will be parallel to the line that passes by the third remaining edge which is (DB), so (DB)//(IJ) and the distance IJ=1/2 DB , this is a property that they have studied, they just named property with no other name, but they can use it in triangles: '' If ABC is a triangle, M and N are the midpoints of two of his edges, then the line (MN) is parallel to the line that passes through the third edge, and the distance MN is half the measure of the third edge.''
After that i tried proving that EDBJ is a parallelogram ( using only basic parallelogram properties ), but i couldn't find a way.

2) I tried also proving that EDJC is a parallelogram,using the fact that its two diagonals bisect each other which should be in the point I, but we only know that I is the midpoint of [DC], we have no information on I being the midpoint of [EJ], so i got stuck again.

3) i thought of central symmetry of a centre I, but i get the same problem seen in 2) , i also tried using vectors ( my brother has only seen the basic definition of the vector sum such as AB+BC=AC, so no big methods here ), but to no avail.

As for the beautifully marvelous mathematician @HallsofIvy who answered me, i know of the similar triangles approach but i couldn't use it as my brother won't study it until next year, so i only have to use basic geometry properties.

 

[Amicci]

I have just had an idea, i will prove that EDJC is a parallelogram, using a permutation between the parallelogram's properties, here it goes:

• The two lines (AD) and (BC ) are parallel because ABCD is a parallelogram, since (AD) and (DE) are identical and (BC) and (JC) are also identical, then : (DE) and (JC) are parallel
  
  
• The quadrilateral EDJC has two diagonals [DC] and [EJ], they intersect at the point I , which happens to be the midpoint of one of them.
  
  So EDJC is a quadrilateral that has two parallel opposing sides, and its diagonals intersect at the midpoint of one of them, is it enough to say that EDJC is a parallelogram ? If so then we will directly conclude that ED = CJ .

 

[Dr.Peterson]

As for the beautifully marvelous mathematician @HallsofIvy who answered me, i know of the similar triangles approach but i couldn't use it as my brother won't study it until next year, so i only have to use basic geometry properties.

This is not about similar triangles, but ASA congruence.

Please tell us exactly what your brother has studied. I would think the congruence theorems are as basic as you can get. The theorems you mention (about parallelograms, for instance) are typically proved using congruence or similarity.

 

[Amicci]

This is not about similar triangles, but ASA congruence.

Please tell us exactly what your brother has studied. I would think the congruence theorems are as basic as you can get. The theorems you mention (about parallelograms, for instance) are typically proved using congruence or similarity.

I am sorry if don't use the proper name, i try to but we study sciences in French and some theorems' names are different. He hasn't studied any of the Triangle Congruence Theorems ( SSS, SAS, ans ASA postulates ), nor did he study the Basic Proportionality Theorem or its converse.
As i mentioned above, i am limited to only use the parallelogram's properties, or central symmetry, or vectors sum, but this is all for early middle school, there are many theorems i couldn't use because they are meant for higher levels. I am beginning to wonder if an information is missing from the exercise's text, as it is meant for early middle school students.

 

[lev888]

After that i tried proving that EDBJ is a parallelogram ( using only basic parallelogram properties ), but i couldn't find a way.

Why use properties? By definition, it's a parallelogram if the opposite sides are parallel. ED || BJ because the original shape is a parallelogram. And you proved that BD || EJ.

 

[Amicci]

Why use properties? By definition, it's a parallelogram if the opposite sides are parallel. ED || BJ because the original shape is a parallelogram. And you proved that BD || EJ.

Love you ! So simple yet not easy to perceive.