A square sheet of paper ABCD is folded with D falling on BC at D', with A falling on A', and A'D' intersecting AB at E.
Prove that the inradius of triangle EBD' is equal to |A'E|.
Let F be the intersection of the fold and AB.
Let r and q be the inradii of the similar right triangles EA'F and EBD' respectively.
r(|EF| - |A'E|) = r|EF| - r|A'E|
= q|ED'| - q|BE|
= q(|A'D'| - |A'E|) - q(|BA| - |EF| - |FA|)
= q(|A'D'| - |A'E|) - q(|A'D'| - |EF| - |FA'|)
= q(|EF| + |FA'| - |A'E|)
= ½(|FA'| + |A'E| - |EF|)(|FA'| - |A'E| + |EF|)
= ½(|FA'|2 - |A'E|^2 + 2|A'E||EF| - |EF|^
= ½(|FA'|2 - |A'E|^2 + 2|A'E||EF| - |FA'|^2 - |A'E|^|2]
= ½(2|A'E||EF| - 2|A'E|^2)
= |A'E|(|EF| - |A'E|)
===> r = |AE|
Edited on February 24, 2024, 12:18 am
Puzzle Thoughts
