The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$.

Question

The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$. Prove that the medians drawn from $A$ and $B$ are perpendicular.

What I Tried: Here is a picture :-

I would normally solve these kind of problems using elementary geometry, but I wasn't able to do it this time, simply because $(a,b,c)$ can take many values and there is probably some kind of construction or some clever proof which I am missing. One idea I thought of is that the medians divide the side lengths into $2$ equal parts, and the centroid itself divides the medians in $2:1$ ratio, and I probably have to show somehow, that Pythagoras Theorem on either of the $3$ triangles work. I took an example in Geogebra by drawing an accurate diagram with $a = 12$ , $b = 18.66$ nearly, and then $c = 10$ . This actually showed that the medians from $A$ and $B$ are really perpendicular, and I just drew a picture of that taken from Geogebra.

Now given $(a,b,c)$ , is there any way to prove this? I did not find any elementary methods like angle-chasing , pythagoras theorem , similarity and areas , etc. , but I did not find any clever use of them till now. Maybe some extra construction is required?

Can anyone help me? Thank You.

Answer 1

Let DB and EA meet at X. From the triangle median theorem $$DB^2= \frac12( AB^2+ BC^2)-\frac14 AC^2, \>\>\> EA^2= \frac12( AB^2+ AC^2)-\frac14 BC^2 $$ Combine them, along with the given $BC^2+AC^2=5AB^2$, to get $$DB^2 +EA^2 = AB^2 +\frac14(BC^2+AC^2)= \frac94 AB^2$$

and substitute it into the cosine rule below \begin{align} \cos\angle DXE= \frac{DX^2+EX^2- DE^2}{2 DX\cdot EX}= \frac{\frac19 (DB^2+EA^2)- \frac14 AB^2}{2DX\cdot EX}=0 \end{align}Thus, $\angle DXE =90^\circ$ and the two medians are perpendicular.

Answer 2

Given four points $W,X,Y,Z$, let $P_{WXYZ} = WX^2 - XY^2 + YZ^2 - ZW^2$. Then we can prove that:

Claim 1. $P_{WXYZ}$ is affine-linear in each of the four points. (That is, as one of the points, say, $X$, moves along a straight line at a constant rate, the value of $P_{WXYZ}$ changes linearly.)

Claim 2. When $W,X,Y,Z$ are all distinct, $P_{WXYZ}=0$ if and only if $WY \perp XZ$.

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So let's compute $P_{ADEB}$. By Claim 1 applied to point $D$ (which is halfway between $A$ and $C$), $P_{ADEB} = \frac12 P_{AAEB} + \frac12 P_{ACEB}$. By Claim 1 applied to point $E$ (which is halfway between $B$ and $C$), we get $$ P_{ADEB} = \frac14 P_{AABB} + \frac14 P_{AACB} + \frac14 P_{ACBB} + \frac14 P_{ACCB}. $$ Since we only want to know if $P_{ADEB}=0$, we can forget about the $\frac14$'s and only expand out these four expressions in terms of $AB^2, BC^2, CA^2$.

We get $$ (AA^2-AB^2+BB^2-BA^2) + (AA^2-AC^2+CB^2-BA^2) + (AC^2-CB^2+BB^2-BA^2) + (AC^2-CC^2+CB^2-BA^2) $$ which simplifies to $AC^2 + BC^2 - 5AB^2$. We are given that this is $0$, therefore $P_{ADEB}=0$ and $AE \perp BD$ by Claim 2.

Answer 3

Note $P$ the intersection point and consider the triangle $\triangle{ABP}$ so we have by the well known formulas for the medians $$AP=\frac23\left(\frac{\sqrt{2(b^2+c^2)-a^2}}{2}\right)\\PB=\frac23\left(\frac{\sqrt{2(a^2+c^2)-b^2}}{2}\right)\\AB=c$$ It is verified directly and easily that $AP^2+PB^2=AB^2$ because of $a^2+b^2=5c^2$