2019 AMC 10B Problem 16

Let $\angle A = \alpha$, so $\angle B = 90^\circ - \alpha$. Since $AC = CD$, $\angle CDA = \alpha$; since $DE = EB$, $\angle EDB = \angle B = 90^\circ - \alpha$. The three angles at $D$ along line $AB$ sum to $180^\circ$, so
$$\angle CDE = 180^\circ - \alpha - (90^\circ - \alpha) = 90^\circ.$$

Scale so $AC = CD = 4$ and $DE = EB = 3$. Right triangle $CDE$ has hypotenuse $CE = 5$, so $BC = CE + EB = 8$ and $AB = \sqrt{4^2 + 8^2} = 4\sqrt5$.

Drop the perpendicular from $C$ to $AB$ at $M$. In isosceles triangle $ACD$, $M$ is the midpoint of $AD$. Right triangles $AMC$ and $ACB$ are similar (they share $\angle A$), so $\frac{AM}{AC} = \frac{AC}{AB}$, giving $AM = \frac{16}{4\sqrt5} = \frac{4}{\sqrt5}$ and $AD = \frac{8}{\sqrt5}$.

Similarly, the perpendicular from $E$ to $AB$ meets it at the midpoint $N$ of $DB$, and $\frac{BN}{EB} = \frac{BC}{AB}$ gives $BN = 3 \cdot \frac{8}{4\sqrt5} = \frac{6}{\sqrt5}$, so $DB = \frac{12}{\sqrt5}$.

Check: $AD + DB = \frac{20}{\sqrt5} = 4\sqrt5 = AB$. Then $AD : DB = 8 : 12 = 2 : 3$.

The answer is $\boxed{\textbf{(A)}\ 2:3}$.

Two isosceles triangles sharing the hypotenuse of a right triangle copy the acute angles $\alpha$ and $90^\circ - \alpha$ to $D$, leaving a right angle between them. That unlocks a Pythagorean computation of every length. Projecting the isosceles apexes onto $AB$ turns each base into twice a leg times a cosine, which similar triangles supply without trigonometry.

With $\cos A = \frac{AC}{AB} = \frac{1}{\sqrt5}$ and $\cos B = \frac{BC}{AB} = \frac{2}{\sqrt5}$, the isosceles bases are $AD = 2 \cdot 4 \cos A = \frac{8}{\sqrt5}$ and $DB = 2 \cdot 3 \cos B = \frac{12}{\sqrt5}$, ratio $2 : 3$.

• Guessing 1:1 because AC = CD and DE = EB look symmetric, or setting AD:DB = AC:DE = 4:3 as if it were an angle bisector.
• Missing that $\angle CDE = 90^\circ$ and trying to coordinate-bash with an unknown leg $BC$, which is far slower.