该图显示了一个正方形 ABCD。 正方形的对角线相交于 W，X 是对角线 BD 上的一个点。Y 位于 CX 上，使得 DY 垂直于 CX 且 DY 与 AC 相交于 Z。

1. Show one pair of corresponding angles are equal right angles.

In a square, the diagonals are perpendicular.

Since $DW$ lies on diagonal $BD$ and $WZ$ lies on diagonal $AC$,

$\angle DWZ = 90^\circ$.

Also, $DY \perp CX$ is given. Since $YZ$ lies on $DY$ and $CY$ lies on $CX$,

$\angle CYZ = 90^\circ$.

Hence,

$\angle DWZ = \angle CYZ$.

2. Show a second pair of corresponding angles are equal and conclude similarity.

At $Z$, the straight lines $AC$ and $DY$ intersect.

Therefore,

$\angle DZW = \angle CZY \quad \text{(vertically opposite angles)}$.

Since two pairs of corresponding angles are equal,

$\triangle DWZ \sim \triangle CYZ \quad \text{(AA similarity)}$.

1. Write down a valid pair of similar but not congruent triangles.

Since $WX$ lies on $BD$ and $WZ$ lies on $AC$, $\angle XWZ=90^\circ$.

Also, $ZY$ lies on $DY$ and $YX$ lies on $CX$, with $DY\perp CX$, so $\angle ZYX=90^\circ$.

Thus a valid pair is

$\triangle WXZ$ and $\triangle YZX$.

1. Set up a coordinate system for the square and represent point X on diagonal BD.

Let the square have side length $a$.

Take

$$A(0,0),\quad B(a,0),\quad C(a,a),\quad D(0,a).$$

Since $X$ lies on diagonal $BD$, whose equation is

$$x+y=a,$$

let

$$X=(t,a-t).$$

2. Find the coordinates of Z using perpendicular gradients and the fact that Z lies on AC.

Gradient of $CX$ is

$$\frac{(a-t)-a}{t-a}=\frac{-t}{t-a}=\frac{t}{a-t}.$$

Since $DY \perp CX$, the gradient of $DY$ is

$$-\frac{a-t}{t}.$$

Line $DY$ passes through $D(0,a)$, so

$$y=a-\frac{a-t}{t}x.$$

Since $Z$ lies on diagonal $AC$, $y=x$. Thus

$$x=a-\frac{a-t}{t}x.$$

Multiplying by $t$,

$$tx=at-(a-t)x.$$

So

$$ax=at$$

and hence

$$x=t.$$

Therefore

$$Z=(t,t).$$

3. Compare the two triangle areas using base and perpendicular height.

For $\triangle BCX$, take base $BC=a$.

The perpendicular distance from $X$ to $BC$ is the horizontal distance from $X$ to the line $x=a$:

$$a-t.$$

So

$$\text{Area of }\triangle BCX=\frac12(a)(a-t).$$

For $\triangle CDZ$, take base $CD=a$.

The perpendicular distance from $Z$ to $CD$ is the vertical distance from $Z$ to the line $y=a$:

$$a-t.$$

So

$$\text{Area of }\triangle CDZ=\frac12(a)(a-t).$$

Hence

$$\text{Area of }\triangle BCX=\text{Area of }\triangle CDZ.$$

1. Use the given corresponding length ratio to find the area ratio of the similar triangles.

From part (a), $\triangle DWZ \sim \triangle CYZ$.

Since $YZ$ corresponds to $WZ$,
$\frac{\text{area of }\triangle CYZ}{\text{area of }\triangle DWZ}=\left(\frac{YZ}{WZ}\right)^2$

Given $YZ:WZ=\sqrt{2}:\sqrt{5}$,
$\frac{\text{area of }\triangle CYZ}{\text{area of }\triangle DWZ}=\left(\frac{\sqrt{2}}{\sqrt{5}}\right)^2=\frac{2}{5}.$

1. Find the area of $\triangle DWZ$ and hence the area of $\triangle CDZ$.

From part (d)(i),
$\frac{\text{area of }\triangle CYZ}{\text{area of }\triangle DWZ}=\frac{2}{5}.$
Given $\text{area of }\triangle CYZ=3\text{ cm}^2$,
$\frac{3}{\text{area of }\triangle DWZ}=\frac{2}{5}$
$\text{area of }\triangle DWZ=\frac{3\times 5}{2}=7.5\text{ cm}^2.$

Since $3CZ=2CW$,
$CZ:CW=2:3.$
As $Z$ lies on $CW$,
$CZ:ZW=2:1.$
Triangles $CDZ$ and $DWZ$ have bases $CZ$ and $ZW$ on the same straight line $AC$, and the same perpendicular height from $D$ to $AC$. Hence
$\frac{\text{area of }\triangle CDZ}{\text{area of }\triangle DWZ}=\frac{CZ}{ZW}=2.$
So
$\text{area of }\triangle CDZ=2(7.5)=15\text{ cm}^2.$

2. Relate $\triangle CDZ$ to the square and calculate the square area.

Since $CZ:CW=2:3$, triangles $CDZ$ and $CDW$ have the same perpendicular height from $D$ to $AC$, so
$\frac{\text{area of }\triangle CDZ}{\text{area of }\triangle CDW}=\frac{CZ}{CW}=\frac{2}{3}.$
Thus
$\text{area of }\triangle CDW=15\times \frac{3}{2}=22.5\text{ cm}^2.$

The diagonals of a square divide it into $4$ equal triangles, so
$\text{area of square }ABCD=4(22.5)=90\text{ cm}^2.$