Regular octagon and triangles

Atti0626

New member
We have a regular octagon ABCDEFGH. Point P is inside the octagon. Show that the sum of the areas of the ABP, CDP EFP, GHP triangles are equal to the sum of the areas of the BCP, DEP, FGP, HAP triangles.

I just have no idea how to start, and this a bonus question, so we wont see the solution of it in the lesson, but I am really interested in this problem and want to see how to solve something like this.
Ideas on how to approach this exercise are welcome, or just some facts that I could use to solve this problem.

Last edited:

Dr.Peterson

Elite Member
What theorems have you learned, that this might be meant to use?

But, also, is this exactly what was said? Or did it say that the SUM of the first four areas equals the SUM of the other four?

After looking at a picture, I'd start by considering pairs of opposite triangles. It's a fun problem, once you see it.

• Atti0626

Atti0626

New member
What theorems have you learned, that this might be meant to use?

But, also, is this exactly what was said? Or did it say that the SUM of the first four areas equals the SUM of the other four?

After looking at a picture, I'd start by considering pairs of opposite triangles. It's a fun problem, once you see it.
I haven't learned theorems in connection of this exercise, we always get bonus problems unrelated to our current topic.
You are right, it is the sum, it was lost when I translated it in English.
Thanks for the tip, I am currently away from home, but I will try to solve it with your idea as soon as I can.

lev888

Full Member
It may help to try with a square first.

• Atti0626

Atti0626

New member
I think I figured it out. If we look at opposing triangles, their base have the same lenght, and the sum of their height is the same in every pair, so every pair's area must be the same, and the equation has 2-2 pairs on both sides, so they must be equal.

• lev888

Dr.Peterson

Elite Member
In other words, each pair's total area is 1/4 of the octagon's area, so the two sums of four triangles are each equal to half the octagon's area.