Three longest edges of a quadrilateral and its diagonals

In our recent class of math history, we discussed many questions about Euclidean geometry. For example, Euclid’s parallel postulate is a postulate that we can use to determine whether two lines are parallel to each other. This postulate somehow reminded me of one challenging geometry question from high school that I did not solve it at that time: In any quadrilateral, is it always true that the sum of the lengths of the quadrilateral’s three longest edges is greater than the sum of the lengths of its two diagonals? The answer is yes. In this blog, I will explore a proof to prove this conclusion.

Before proving our original question, let’s first prove a lemma. Let the following figure be a reference:

In triangle ABC, if angle C is greater than or equal to 90 degrees, then we have

AB + CH > AC + BC.

Let’s assume the angle of C is 90 degrees. According to the Pythagorean theorem, we have

AB * AB = AC * AC + BC * BC (1)

and we have two ways to calculate the area of the triangle ABC. So we can have

AB * CH = AC * BC (2).

figure1

Some people may be wondering how to prove the question by using this lemma? Let’s keep reading and see what the connection is.

Now consider the left side of (1) plus two times of the left side of (2) and let the right side of (2) plus two times of the right side of (2), which creates

AB * AB + 2 * AB * CH = AC * AC + 2 * AC * BC + BC * BC.

So we can say

AB * AB + 2 * AB * CH + CH * CH is greater than AC * AC + 2 * AC * BC + BC * BC.

Now we can simplify the equality into

(AB + CH)^2 > (AC + BC)^2.

Now take the square root on both sides and we have

AB + CH > AC + BC.

What about if the angle C is greater than 90 degrees? Will our proof still be true? Now let’s draw a line CD so that the angle of ACD is 90 degrees. In the above paragraph, we already proved that AD + CH > AC + DC (2).

Because of the truth that the sum of two triangle edges must be greater than the third edge, we have BD + DC > BC (2).

Let us combine inequality (1) and (2), creating

figure2

AD + CH + BD + DC > AC + DC + BC.

From the above figure, we know

AD + BD = AB

so

AB + CH + DC> AC + BC + DC.

Now subtract DC from both sides and we will have

AB + CH > AC + BC.

Therefore, we proved our lemma.

Now it is time to see how we can get the answer for our original question. Let’s use the following figure as a reference. Let’s make two parallelograms ABDE and BCFD. Since ABDE is a parallelogram,

figure3

Image: Jinggong Zheng.

we have

AE is parallel to BD and AE = BD.

BD is parallel to CF and BD = CF

because BCFD is a parallelogram.

Therefore,

AE and CF are parallel to each other and AE = CF.

So we can conclude that quadrilateral ACFE is also a parallelogram. In parallelogram ACFE,

the sum of DA, DE, DC and DF = the sum of AB, BC, CD and AD

and AE = CF = BD and EF = AC.

In order to prove our original question, we just need to show that

any of three sums of DA, DC, DE, and DF > sum of two edges in ACFE.

We know that the two opposite angles in a parallelogram must be equal. So one of the opposite angles in ACFE must be greater than or equal to 90 degrees. For example, in the above figure, see angle AEF and ACF.

Let’s draw a line AF and this parallelogram will be divided into two triangles ACF and AEF. Let’s make a line CH in ACF.

According to our previous lemma, we must have

AF + CH > AC + FC.

Any sum of the two edges in the triangle ADF will be greater than the third edge, therefore we have

AD + DF > AF.

H is the point that will give us the shortest distance from C and D is on the other side of AF, therefore CD > CH. Finally, we have

AD + DF + CD > AF + CH > AC + FC.

This is an interesting proof, right? We first proved a simple lemma and then used that lemma to prove our original question. If we keep think about the same question in hyperbolic or spherical, will this proof still work for us? My personal assumption is: No. The reason is, the Pythagorean theorem will not hold in a non-Euclidean geometry. Therefore, we cannot prove the lemma as above.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s