2019 AMC 10B Problem 16

Question

In

\triangle ABC

with a right angle at

C,

point

D

lies in the interior of

\overline{AB}

and point

E

lies in the interior of

\overline{BC}

so that

AC=CD,

DE=EB,

and the ratio

AC:DE=4:3.

What is the ratio

AD:DB?

2:3

2:\sqrt{5}

1:1

3:\sqrt{5}

3:2

Scale so

AC=CD=4

and

DE=EB=3

Use isosceles angles to show

\angle EDC=90^\circ

Scale the figure so that

AC=CD=4

and

DE=EB=3

. Let

A=\angle BAC

and

B=\angle ABC

, so

A+B=90^\circ

.

The isosceles triangles

ACD

and

DEB

give

\angle CDA=A

and

\angle EDB=B

. Because

DA

and

DB

are opposite rays,

\angle CDE=180^\circ-A-B=90^\circ.

Therefore

CE=\sqrt{4^2+3^2}=5

, so

BC=CE+EB=8

and

\tan A=\frac{BC}{AC}=2

.

The bases of the two isosceles triangles have lengths

AD=8\cos A

and

BD=6\cos B=6\sin A

. Hence

\frac{AD}{BD}=\frac{8\cos A}{6\sin A}=\frac{4}{3\tan A}=\frac23.

Thus, the answer is A .

Problem 16 in other years