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 Another triangle in a square problem (Posted on 2023-03-29)
ABCD is a square with point E inside the square.
Line segments AE, BE, and DE are drawn.
The length of AE is 8 and angle AEB is a right angle.

What is the area of triangle AED?

 Submitted by Brian Smith No Rating Solution: (Hide) 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

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