When do two triangles reflected over midpoints have the same area?

Question

Suppose I have a triangle ABC. I have points C', A', and B' on segments AB, BC, and CA, respectively. Suppose I reflect C' about the midpoint of AB to get point C'' (also on AB); similarly for the other two points to get points A'' and B''.

Is the area of A'B'C' the same as the area of A''B''C''?

I tried some programming and it looks like it is. How can I prove this? I tried proving this showing that Area(AC'B')+Area(BA'C')+Area(CA'B') = Area(AC''B'')+Area(BA''C'')+Area(CA''B''). I did this by saying that AB has length c, BC has length A, and CA has length b.

Then I decided to show that the sum of the areas of the three outside triangles of each triangle is the same. For instance the area of AC'B' is $1/2 (1/2 c - \delta_{B'})(1/2 b - \delta_{C'})\sin A$, where $\delta_{B'}$ is the distance of $B'$ from the midpoint of AC. I wrote out all of the areas of the triangles like this, but I cannot show that they are the same.

Answer 1

Here is a way to prove it. I will not go into details, it would take long to type it in Latex, but you will easily understand the idea.

Use the determinant formula for the area of a triangle using coordinates: namely, the first column are the $x$-coordinates of the vertices, the the second column are the $y$-coordinates of the vertices, and the third column is all ones (well, you have to include a factor of $1/2$). The points $A'$, $B'$ and $C'$ have the form $tB+(1-t)C$, $sA+(1-s)C$ and $rA+(1-r)B$ for some $t,s,r$. Use the determinant formula to express the area of $A'B'C'$ as a determinant.

The symmetric points are $A''=(1-t)B+tC$, $B''=(1-s)A+sC$ and $C''=(1-r)A+rB$. Use the determinant formula to express the area of $A''B''C''$ as a determinant.

Now use the multi-linearity of the determinant to prove that the determinants are equal (each of your determinants equals the sum of four determinants, you can prove that they are equal using basic properties).

Note that the proof works for negative $t,r,s$ so the points do not have to be strictly inside the sides, as long as they are symmetrical w.r.t. the midpoint.

Answer 2

Analytic geometry(On the coordinate plane)

Let us say that $B=(0,0),A=(a,b)$ and $C=(c,0)$.

Let $C'=(t,\frac {b}{a}t)$, $A'=(q,0)$, $B'=(p,\frac{bp-bc}{a-c})$

(points $A',B',C'$can be defined as above by calculating the gradient of $AB,BC,CA$.)

By calculating the symmetrical points of $A',B',C'$ w.r.t the midpoints of $BC,CA,AB$, We can say that

$C''=(a-t,b-\frac{b}{a}t), A''=(c-q,0), B''=(a+c-p,\frac{ab-bp}{a-c})$.

According to the shoelace formula,

One could "calculate" the Area of triangle $A'B'C'$ and $A''B''C''$.

So the problem is true.