A B C D is a square with side length of 1. △ B E F inscribed in the square is such that E is on A D and F is on C D . Perimeter of △ D E F is 2. Find the measure of ∠ E B F .
Bonus : Use geometry only.
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If we draw a circle with center at B and radius equal to |AB| = 1, we know that the perimeter of any triangle DEF, which has got EF tangent to the circle is equal to 2, because |AE|=|EG| and |FG|=|FC| so the perimeter of triangle DEF is equal to |AD| + |CD| = 2. Now to prove that these triangles DEF are the only ones with perimeter 2 we can consider this: Draw any other line segment E´F´ and we can for every E´F´ find paralel line which is tangent to the quarter circle. Because
E´F´ and the new tangent lines are paralel we can say that triangles DE´F´ and a new triangle made by the paralel line are similar with coeficient of similarity not equal to 1 (because line E´F´ is not same as the new line) so the perimeter is not equal to 2. Now we can easly calculate angle EBF as a half of the sum of angles ABG and GBC so it is 45°.
Let D F = x and D E = y . Since the perimeter of △ D E F is 2,
x + y + x 2 + y 2 x + y − 2 x 2 + 2 x y + y 2 − 4 ( x + y ) + 4 ⟹ x y = 2 = x 2 + y 2 = x 2 + y 2 = 2 ( x + y − 1 ) Squaring both sides
Let ∠ E B F = θ . Then
θ = 9 0 ∘ − tan − 1 ( 1 − x ) − tan − 1 ( 1 − y ) = 9 0 ∘ − tan − 1 ( 1 − ( 1 − x ) ( 1 − y ) 1 − x + 1 − y ) = 9 0 ∘ − tan − 1 ( x + y − x y 2 − x − y ) = 9 0 ∘ − tan − 1 ( x + y − x y 2 − x − y ) = 9 0 ∘ − tan − 1 ( 2 − x − y 2 − x − y ) = 9 0 ∘ − tan − 1 ( 1 ) = 9 0 ∘ − 4 5 ∘ = = 4 5 ∘ Note that x y = 2 ( x + y − 1 )
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