Hexagon Area Question

Question

Triangle ABC has AB = 15, BC = 13, and AC = 14. Let O be the orthocenter of this triangle, and let the reflections from the orthocenter across sides AB, BC, and AC be points D, E, and F, respectively. Find the area of ADBECF.

I was able to show that this hexagon was concyclic, but I don't know the best way to proceed from there.

By the reflection construction, $\triangle AOB \cong \triangle ADB$ etc. The areas of $\triangle AOB, \triangle BOC, \triangle COA$ together is the area of $\triangle ABC$. — peterwhy, Nov 17 '23 at 19:59

KDP · Answer 1 · 2023-11-19T20:07:07.153

3

The area is double the area of the initial triangle. The triangle is divided into 3 sections and reflecting each of those sections outside the triangle creates a shape with twice the area. The area of the initial triangle can be calculated using just the known side lengths of the triangle and Heron's formula

$$a = 2\sqrt{p(p-13)(p-14)(p-15)} = 168$$ where the semi-perimeter $p = (13+14+15)/2$

edited Nov 19 '23 at 20:07

answered Nov 19 '23 at 18:15

KDP

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Hexagon Area Question

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