Subfield: Geometry


Question: In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$.

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Solution:
Rotate $\triangle D F C$ through an angle of $90^{\circ}$ counterclockwise about $D$, so that $D C$ now lies along $D A$ and $F^{\prime}$ is outside the square, as shown.

Join $F^{\prime}$ to $E$.

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Since $A C$ is a diagonal of square $A B C D$, then $\angle E A D=\angle F C D=45^{\circ}$.

Since $\angle E A D=45^{\circ}$ and $\angle F^{\prime} A D=\angle F C D=45^{\circ}$, then $\angle F^{\prime} A E=45^{\circ}+45^{\circ}=90^{\circ}$.

By the Pythagorean Theorem in $\triangle F^{\prime} A E$, we have $F^{\prime} E^{2}=F^{\prime} A^{2}+A E^{2}$.

But $F^{\prime} A=F C=z$ and $A E=x$, so $F^{\prime} E^{2}=z^{2}+x^{2}$.

To show that $y^{2}=x^{2}+z^{2}$, it is sufficient to show that $F^{\prime} E=y$.

Consider $\triangle F^{\prime} D E$ and $\triangle F D E$.

Note that $F^{\prime} D=F D$ and $\angle F^{\prime} D A=\angle F D C$ by construction and $E D=E D$.

Also, $\angle F^{\prime} D E=\angle F^{\prime} D A+\angle E D A=\angle F D C+\angle E D A=90^{\circ}-\angle E D F=45^{\circ}$, which tells us that $\angle F^{\prime} D E=\angle F D E=45^{\circ}$.

Therefore, $\triangle F^{\prime} D E$ is congruent to $\triangle F D E$ (side-angle-side), and so $F^{\prime} E=F E=y$.

This gives us the desired conclusion that $y^{2}=x^{2}+z^{2}$.
Since $A C$ is a diagonal of square $A B C D$, then $\angle E A D=\angle F C D=45^{\circ}$.

Let $\angle A D E=\theta$.

Since the angles in a triangle have a sum of $180^{\circ}$, then

$$
\angle A E D=180^{\circ}-\angle E A D-\angle A D E=180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta
$$

Since $A E F$ is a straight angle, then $\angle D E F=180^{\circ}-\angle A E D=180^{\circ}-\left(135^{\circ}-\theta\right)=45^{\circ}+\theta$. Continuing in this way, we find that $\angle E F D=90^{\circ}-\theta, \angle D F C=90^{\circ}+\theta$, and $\angle F D C=45^{\circ}-\theta$.

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Using the sine law in $\triangle A E D$, we see that $\frac{A E}{\sin \angle A D E}=\frac{E D}{\sin \angle E A D}$ or $\frac{x}{\sin \theta}=\frac{E D}{\sin 45^{\circ}}$.

Using the sine law in $\triangle D E F$, we see that $\frac{E F}{\sin \angle E D F}=\frac{E D}{\sin \angle E F D}$ or $\frac{y}{\sin 45^{\circ}}=\frac{E D}{\sin \left(90^{\circ}-\theta\right)}$.



Using the sine law in $\triangle D E F$, we see that $\frac{E F}{\sin \angle E D F}=\frac{F D}{\sin \angle D E F}$ or $\frac{y}{\sin 45^{\circ}}=\frac{F D}{\sin \left(45^{\circ}+\theta\right)}$.

Using the sine law in $\triangle D F C$, we get $\frac{F C}{\sin \angle F D C}=\frac{F D}{\sin \angle D C F}$ or $\frac{z}{\sin \left(45^{\circ}-\theta\right)}=\frac{F D}{\sin 45^{\circ}}$.

Dividing the first of these equations by the second, we obtain $\frac{x \sin 45^{\circ}}{y \sin \theta}=\frac{\sin \left(90^{\circ}-\theta\right)}{\sin 45^{\circ}}$ or $\frac{x}{y}=\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\sin ^{2} 45^{\circ}}$.

Dividing the fourth of these equations by the third, we obtain $\frac{z \sin 45^{\circ}}{y \sin \left(45^{\circ}-\theta\right)}=\frac{\sin \left(45^{\circ}+\theta\right)}{\sin 45^{\circ}}$ or $\frac{z}{y}=\frac{\sin \left(45^{\circ}+\theta\right) \sin \left(45^{\circ}-\theta\right)}{\sin ^{2} 45^{\circ}}$.

Since $\sin \left(90^{\circ}-\alpha\right)=\cos \alpha$ for every angle $\alpha$, then $\sin \left(90^{\circ}-\theta\right)=\cos \theta$.

Also, $\sin \left(45^{\circ}+\theta\right)=\sin \left(90^{\circ}-\left(45^{\circ}-\theta\right)\right)=\cos \left(45^{\circ}-\theta\right)$.

Using this and the fact that $\frac{1}{\sin ^{2} 45^{\circ}}=\frac{1}{(1 / \sqrt{2})^{2}}=2$, the expressions for $\frac{x}{y}$ and $\frac{z}{y}$ become

$$
\frac{x}{y}=2 \cos \theta \sin \theta=\sin 2 \theta
$$

and

$$
\frac{z}{y}=2 \cos \left(45^{\circ}-\theta\right) \sin \left(45^{\circ}-\theta\right)=\sin \left(2\left(45^{\circ}-\theta\right)\right)=\sin \left(90^{\circ}-2 \theta\right)=\cos 2 \theta
$$

Finally, this tells us that

$$
\frac{x^{2}}{y^{2}}+\frac{z^{2}}{y^{2}}=\left(\frac{x}{y}\right)^{2}+\left(\frac{z}{y}\right)^{2}=\sin ^{2} 2 \theta+\cos ^{2} 2 \theta=1
$$

Since $\frac{x^{2}}{y^{2}}+\frac{z^{2}}{y^{2}}=1$, then $x^{2}+z^{2}=y^{2}$, as required.




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