2
$\begingroup$

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 enter image description here

$\endgroup$
1
  • $\begingroup$ What have you tried? $\quad$ Are you familiar with setting them up as complex numbers or vectors, to calculate the length and angle? Honestly, even brute force coordinate geometry could work. $\endgroup$ Commented May 21, 2024 at 15:20

2 Answers 2

Reset to default
3
$\begingroup$

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.
$\endgroup$
2
  • $\begingroup$ Thank you all so much for your help and answers! The problem was beautifully solved using geometric reflections and midpoint properties, and I'm very satisfied with the result. $\endgroup$ Commented May 22, 2024 at 4:42
  • $\begingroup$ @OthS Please upvote and accept solutions that you greatly appreciate. Thanks. $\endgroup$ Commented May 22, 2024 at 5:01
1
$\begingroup$

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 $w, z$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.