Is a proof "The diagonals of a parallelogram bisect each other" without using concept of "congruence of angles" still correct?

Question

I post my proof in pictures below and I am not sure if my proof correct?

This is an exercise from section 1.1 of the classical textbook linear algebra by Stephen Friedberg etc

Prove that the diagonals of a parallelogram bisect each other.

I choose to use words and ideas from this section as possible as I can to prove that exercise.

And surfing on the internet I found that other ways of this proof use concept of "congruence of angles"

Is a proof "The diagonals of a parallelogram bisect each other" without using concept of "congruence of angles" still correct?

Answer 1

My idea is that just using PARALLELOGRAM LAW FOR VECTOR ADDITION

then two diagonals have the same midpoint.

So the conclusion must be that the diagonals of a parallelogram bisect each other.

https://i.stack.imgur.com/mcAWX.jpg

Answer 2

Here's a proof that doesn't use angle congruences, using vectors instead.

Let $\vec a$ and $\vec b$ span the parallelogram.

The diagonal between the points $a$ and $b$ is described by $t\vec a + (1-t)\vec b,\quad 0\le t\le 1$, and it's midpoint at $t=\frac12$ is also the midpoint of the other diagonal by symmetry.

Answer 3

Here's a standard proof based on my experience with vectors.

Let $\ WXYZ\ $ be the vertices of a paralellogram, in clockwise order, and let $\ P\ $ be the point of intersection of the diagonals $\ WY\ $ and $\ XZ.$

$\ \exists\ \lambda\in\mathbb{R}:\ \overset{\longrightarrow}{WP} = \lambda \overset{\longrightarrow}{WY} = \lambda\left( \overset{\longrightarrow}{WX} + \overset{\longrightarrow}{XY}\right)\qquad (1) $

$\ \exists\ \mu\in\mathbb{R}:\ \overset{\longrightarrow}{XP} = \mu \overset{\longrightarrow}{XZ} = \mu\left( \overset{\longrightarrow}{XY} + \overset{\longrightarrow}{YZ}\right) \implies \overset{\longrightarrow}{WP} = \overset{\longrightarrow}{WX} + \overset{\longrightarrow}{XP} = \overset{\longrightarrow}{WX} + \mu\left( \overset{\longrightarrow}{XY} + \overset{\longrightarrow}{YZ}\right) = \overset{\longrightarrow}{WX} + \mu\left( \overset{\longrightarrow}{XY} - \overset{\longrightarrow}{WX}\right) = (1 - \mu) \overset{\longrightarrow}{WX} + \mu \overset{\longrightarrow}{XY} \qquad (2) $

$(1)$ and $(2)$ then give:

$$ (1 - \mu) \overset{\longrightarrow}{WX} + \mu \overset{\longrightarrow}{XY} = \overset{\longrightarrow}{WP} = \lambda\ \overset{\longrightarrow}{WX} + \lambda\ \overset{\longrightarrow}{XY},$$

and since $\ \overset{\longrightarrow}{XY}\ $ and $\ \overset{\longrightarrow}{WX}\ $ are not parallel, we may equate coefficients of $\ \mu\ $ and $\ \lambda\ $ to get that $\ \lambda = \mu = \frac{1}{2},\ $ and so $\ \overset{\longrightarrow}{WP} = \frac{1}{2}\left( \overset{\longrightarrow}{WX} + \ \overset{\longrightarrow}{XY} \right)\ = \frac{1}{2} \overset{\longrightarrow}{WY},\ $ and I'll leave it to you to show that therefore $\ \overset{\longrightarrow}{XP} = \frac{1}{2}\overset{\longrightarrow}{XZ}.$