Another triangle in a square problem (Posted on 2023-03-29)

	Submitted by Brian Smith
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There are a few trigonometric/calculus solutions in the comments, but here I present a purely geometric solution. Draw the line perpendicular to AE which passes through D. (If angle AED is obtuse then extend AE as needed.) Call the intersection point F. Angle ABE and angle BAE are complementary from being the acute angles of right triangle AEB. Angle BAE and angle DAE (same as DAF) are complementary from being two halves of right angle BAD. Thus angle BAE and angle DAE are congruent. Angle BAE and angle DAF are congruent; angle AEB and angle AFD are right angles; and AB=AD. Thus triangles AEB and DAF are congruent right triangles. Then from congruent triangles AEB and DAF we must have DF=AE=8. DF is also the altitude of triangle AED corresponding to side AE. Then the area of triangle AED is DF*AE/2 = 8*8/2 = 32.

	Subject	Author	Date
	my solution	Charlie	2023-03-29 21:52:20
	Puzzle Thoughts	K Sengupta	2023-03-29 21:27:37
	Simple trig solution	Jer	2023-03-29 14:33:56
	Solution	Larry	2023-03-29 12:02:49