The altitude from the right angle to the hypotenuse creates two smaller right triangles that are each similar to the original triangle. Think about writing a proportion involving $DE$, $DF$, and $DG$, and another involving $EF$, $DF$, and $GF$.

For angle $D$ in the original triangle, which leg is opposite and which is adjacent? Express $\tan(D)$ as a ratio of these legs, and consider how your equations for $DE^2$ and $EF^2$ can help you find this ratio without directly taking square roots at first.

In Desmos, type 54/6 and note the value. This corresponds to the ratio $GF/DG$ that appears when you form $\dfrac{EF^2}{DE^2}$.

In Desmos, type sqrt(54/6) (or sqrt(ans) if your calculator supports using the previous answer). The displayed value is the value of $\tan(D)$.

• $\angle E$ is the right angle, so $\overline{DF}$ is the hypotenuse of $\triangle DEF$.
• The altitude from $E$ meets $\overline{DF}$ at $G$, splitting the hypotenuse into two parts: $DG = 6$ and $GF = 54$.
• So the whole hypotenuse is

Dropping an altitude from the right angle in a right triangle creates two smaller right triangles that are similar to the original triangle.

• $\triangle DEF$ and $\triangle DEG$ are both right triangles and share angle $D$, so they are similar.
• From similarity, the ratio of a leg to the hypotenuse in the big triangle equals the ratio of the corresponding leg to the hypotenuse in the smaller triangle:

• Cross-multiply to get a relationship involving $DE^2$:

Now use the other small triangle.

• $\triangle DEF$ and $\triangle EGF$ are also similar (both are right triangles and share angle $F$).
• So we have a similar proportion for $EF$:

• Cross-multiply to get a relationship involving $EF^2$:

By definition in a right triangle,

• For angle $D$, the opposite side is $EF$.
• The adjacent side is $DE$.

So

To use the equations for $DE^2$ and $EF^2$, square the tangent:

Substitute the expressions we found:

Since angle $D$ is acute, $\tan(D)$ is positive, so

The correct answer is $3$.