2019 AMC 10B Problem 16

We know angles more easily than lengths here, so start with angles at $D$.

Step 1: angle $CDE$ is a right angle.

1. Let angle $A$ be $\alpha$. Angle $C$ is $90^\circ$, so angle $B$ is $90^\circ - \alpha$.
2. Triangle $ACD$ has $AC = CD$, so its base angles match: angle $CDA = \alpha$.
3. Triangle $DEB$ has $DE = EB$, so angle $EDB =$ angle $B = 90^\circ - \alpha$.
4. The three angles at $D$ along the straight line $AB$ add to $180^\circ$:
$$\angle CDE = 180^\circ - \alpha - (90^\circ - \alpha) = 90^\circ.$$

Step 2: find the legs of the big triangle.

1. Only the ratio matters, so take $AC = CD = 4$ and $DE = EB = 3$.
2. Triangle $CDE$ has a right angle at $D$ and legs $4$ and $3$, so $CE = 5$.
3. Then $BC = CE + EB = 5 + 3 = 8$, and $AC = 4$.

Step 3: locate $D$ with coordinates.

8. Put $C = (0, 0)$, $A = (0, 4)$ and $B = (8, 0)$.
9. Moving from $A$ toward $B$, a point that has gone a fraction $t$ of the way is $(8t,\ 4 - 4t)$.
10. $D$ is the point on $AB$ (other than $A$) at distance $4$ from $C$:
$$(8t)^2 + (4 - 4t)^2 = 16 \;\Longrightarrow\; 64t^2 + 16 - 32t + 16t^2 = 16 \;\Longrightarrow\; 80t^2 = 32t.$$
Since $t \ne 0$, we get $t = \frac{32}{80} = \frac25$.
11. So $D$ is $\frac25$ of the way from $A$ to $B$, and $AD : DB = \frac25 : \frac35 = 2 : 3$.

Check with the other condition: $D = \left(\frac{16}{5}, \frac{12}{5}\right)$ and $E = (5, 0)$, so $DE^2 = \left(\frac95\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{81 + 144}{25} = 9$. So $DE = 3 = EB$. It fits.

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

Isosceles triangles copy angles: equal sides face equal angles. Here the two isosceles triangles move the angles at $A$ and $B$ over to point $D$, and since those two angles add to $90^\circ$, the angle left between them is a right angle. That right angle unlocks a $3$-$4$-$5$ triangle, and then every length follows. When a figure has isosceles triangles sitting on a line, chase the angles at the shared point.

Drop a perpendicular from $C$ to $AB$, meeting it at $M$. In isosceles triangle $ACD$ it cuts the base in half, so $AD = 2 \cdot AM$. Triangles $AMC$ and $ACB$ are similar (both have a right angle and share angle $A$), and $AB = \sqrt{4^2 + 8^2} = 4\sqrt5$, so $AM = \frac{AC \cdot AC}{AB} = \frac{16}{4\sqrt5}$ and $AD = \frac{8}{\sqrt5}$. The same idea from $E$ gives $DB = 2 \cdot \frac{EB \cdot BC}{AB} = \frac{12}{\sqrt5}$. The ratio is $8 : 12 = 2 : 3$.

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