Hello, OnlyMath!

I did a lot of trig and came with a surprisingly simple answer.

I suspect that there is an elegant geometric solution . . . somewhere.

is a triangle where the point is on

. . such that: .

FindCode:B o * * * 20 * * * D a * 100 o * * 80 * a * 60 * x * * * @ * A o * * * * * * * o C

We have: .

Let

.[1]

.[2]

Equate [1] and [2]:

. .

. .

. .

. .

. .

. . . . . . .

Therefore: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Fascinating!

is isosceles: .

Also, is isosceles: .

Is there a more direct way to derive those facts?