Right triangle inside a square

Question

Hello I'm trying to prove the following Let ABCD be a square where X is the midpoint of segment AB and BY = (1/4)BC where DXY is a triangle Prove DXY is a right triangle (using coordinate approach)

Diagram of what im trying to show

Now I'm not quite sure how to approach this problem, My idea on approaching the proble is to show XY and XD have negative reciproals of one another i.e same slope just one is negative and one is positive. Yet I am unsure because of the whole square that encompasses the triangle is supposed to used.

Answer 1

Here is a possible way to solve the problem: Our aim is to prove that $\angle YXD=90^\circ$

Let, the side length of the square be $l$.

As given (See diagram), $AX=BX=\frac{l}{2}$ and $BY=\frac{1}{2}BX=\frac{1}{2}AX$. Let, $\angle BYX=p,\angle BXY=q, \angle AXD=r, \angle ADX=r$.

Then clearly $r+s=90^\circ$. Also $\frac{BY}{BX}=\frac{XA}{AD}=\frac{1}{2}\implies q=s$ and since $\angle A=\angle B=90^\circ$ hence $p=r$. Thus $q+r=90^\circ\implies\angle XYD=90^\circ$. This completes the proof.

Answer 2

Let $A$ be an origin, $B(0,4),$ $C(4,4),$ $D(4,0)$.$

Thus, $X(0,2)$ and $Y(1,4)$ and

$$m_{XY}\cdot m_{XD}=\frac{2-4}{0-1}\cdot\frac{2-0}{0-4}=-1,$$ which says $\measuredangle YXD=90^{\circ}.$

Here $m_{XY}$ is a slope of $XY$.

Answer 3

Let the side of square be $a$ and mark the square ABCD clockwise so that origin be at D, we have:

$(x, y)_X=(a/2, a)$

$(x, y)_Y=(a, 3a/4)$

Then:

$DY^2=a^2+(3a/4)^2=\frac{25a^2}{16}$

$DX^2=(a/2)^2+a^2=\frac{5a^2}{4}$

$XY^2=(a-a/2)^2+(3a/4-a)^2=\frac{5a^2}{16}$

$XY^2+DX^2=\frac{5a^2}{16}+\frac{5a^2}{4}=\frac{25a^2}{16}=DY^2$

So the triangle is right angle.

Answer 4

Your initial approach is valid and is probably the easiest way to solve this problem. Let's compare the slopes of the two lines $XY$ and $XD$. Remember that our slope can be written as $$m = \frac{\triangle y}{\triangle x}$$, where $\triangle y$ is the change in $y$, and $x$ is the change $x$.

Now the slope of $XY = \frac{2}{1} = 2$

Similarly, the slope of $XD = \frac{-2}{4} = \frac{-1}{2}$

Multiplying the slopes together, we see that $2 \cdot \frac{-1}{2} = -1 $, and therefore, our two lines are perpendicular

Quick Note: All calculations made were based on the diagram given above