Proving Perpendicularity and Equal Lengths in Rotated Isosceles Right Triangles

Question

Given two isosceles right triangles $( \triangle ABC )$ and $( \triangle BDE )$ with $( \angle BAC = \angle BDE = 90^\circ )$, triangle $( \triangle BDE )$ is rotated counterclockwise by an angle $( \gamma ) (where ( 0^\circ < \gamma < 360^\circ ))$ about point $( B )$. Point $( E )$ is then connected to point $( C )$, and $( F )$ is the midpoint of segment $( EC )$. It appears that segments $( AF )$ and $( DF )$ are always perpendicular and equal in length regardless of the rotation angle $( \gamma )$.

Here is a diagram to illustrate the problem:

I am looking for a proof of why $( AF \perp DF )$ and $( AF = DF )$ hold true for any rotation angle $( \gamma )$.

What I've considered so far:

1. Since both triangles are isosceles right triangles, their properties might be useful.
2. The rotation of $( \triangle BDE )$ around point $( B )$ does not change the relative properties of points $( D )$ and $( E )$.

I need help with a formal proof to show that $( AF )$ and $( DF )$ are always perpendicular and equal in length regardless of $( \gamma )$. How can we approach this problem? https://www.geogebra.org/classic/eghcku7y

Answer 1

Here's a synthetic solution.

• Let $C'$ be the reflection of $C$ about $A$.
  • Since $F$ is the midpoint of $CE$, and $A$ is the midpoint of $CC'$, hence $ C'E$ is parallel to $AF$ and twice its length.
• Let $E'$ be the reflection of $E$ about $D$.
  • Since $F$ is the midpoint of $CE$, and $D$ is the midpoint of $EE'$, hence $ E'C$ is parallel to $DF$ and twice its length.
• Observe that $BCE'$ and $BC'E$ are $90^\circ$ rotations about $B$ (Check side lengths and angle in between), hence $C'E$ and $E'C$ are of the same length and are perpendicular to each other.
• Hence $AF$ and $DF$ are of the same length (half of $|CE'|=|EC'|$) and perpendicular to each other.

Answer 2

If you're familiar with complex numbers, then this has a quick easy solution.

Let's rephrase the problem in the complex plane.
Let $B = 0$ be the origin.
Let $C = w$ and $ E = z$ be the base of our triangles. We have $ F = \frac{ w + z } { 2}$.
Let $ j = e^{ i \pi / 4 } = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$. Multiplication by $j$ will rotates the complex number by $45^\circ $clockwise, so $D = \frac{\sqrt{2}}{2j} z $ and $ A = \frac{\sqrt{2}}{2} jw$.
The problem becomes

Verify the arithmetic: $FA \times i = FD$.
This implies that segments $FA$ and $FB$ are obtained by a $90^\circ$ rotation, hence are perpendicular and of equal length, regardless of the points $C$ and $E$.

This is a simple verification:
$FA = A - F = \frac{1}{2} [ w (\sqrt{2}j -1) + z(-1)]$
$FD = D - F = \frac{1}{2} [ w (-1) + z(\frac{\sqrt{2}}{j} - 1) ]$.
Since $ (\sqrt{2}j -1) \times i = i \times i = - 1$ and $ (\frac{\sqrt{2}}{j} - 1) = (1-i - 1) = (-1) \times i $, hence we have $ FA \times i = FD$ for all complex numbers $w, z$.