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2009 AMC 10A Problem 17

2009 AMC 10A Problem 17AMC 10 Step by Step

Name: Date: September 13, 2026

Similar & Congruent Triangles · difficulty 4 of 5 · about 6 min · position P16-20

Rectangle $ABCD$ has $AB=4$ and $BC=3$ . Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$ , and $A$ and $C$ lie on $DE$ and $DF$ , respectively. What is $EF$ ?

A)
$9$
B)
$10$
C)
$\frac {125}{12}$
D)
$\frac {103}{9}$
E)
$12$

Answer

Key insight

EF is the hypotenuse of right triangle DEF with altitude DB = 5; triangles DBE and DBF are scaled copies of the 3-4-5 triangles DAB and DCB.

Solution

Draw the rectangle with $D$ at the top left, $A$ below it, and $C$ to its right. The diagonal $DB$ has length $5$ . Since $A$ is on $DE$ , the point $E$ lies on line $DA$ extended beyond $A$ ; since $C$ is on $DF$ , the point $F$ lies on line $DC$ extended beyond $C$ . Thus $\angle EDF = \angle ADC = 90^\circ$ , and $DB$ is the altitude from $D$ to the hypotenuse $EF$ of right triangle $DEF$ .

Look at $\triangle DBE$ : it is right-angled at $B$ and shares $\angle BDA$ with $\triangle DAB$ , which is right-angled at $A$ . So $\triangle DAB \sim \triangle DBE$ with $DA \to DB$ , i.e. scale factor $\frac{DB}{DA} = \frac{5}{3}$ . Hence
$BE = AB \cdot \frac{5}{3} = \frac{20}{3}.$

Similarly $\triangle DCB \sim \triangle DBF$ (right angles at $C$ and $B$ , common angle at $D$ ) with scale factor $\frac{DB}{DC} = \frac{5}{4}$ , so
$BF = CB \cdot \frac{5}{4} = \frac{15}{4}.$

Therefore
$EF = BE + BF = \frac{20}{3} + \frac{15}{4} = \frac{80 + 45}{12} = \frac{125}{12}.$

The answer is $\boxed{\textbf{(C)}\ \frac{125}{12}}$ .

Why this works

A perpendicular to a diagonal at its endpoint creates right triangles that share an acute angle with the rectangle's own right triangles, so everything is a scaled $3$ - $4$ - $5$ . Scaling by $\frac{\text{new hypotenuse-side}}{\text{old}}$ gives each piece of $EF$ in one line. Recognizing that $DB$ is an altitude to the hypotenuse of $\triangle DEF$ also opens the geometric-mean relations $DE = \frac{DB^2}{DA}$ and $DF = \frac{DB^2}{DC}$ .

Alternative approach

Coordinates: $A = (0,0)$ , $B = (4,0)$ , $C = (4,3)$ , $D = (0,3)$ . Line $DB$ has slope $-\frac34$ , so $EF$ through $B$ has slope $\frac43$ : $y = \frac43(x - 4)$ . It meets line $DA$ ( $x = 0$ ) at $E = (0, -\frac{16}{3})$ and line $DC$ ( $y = 3$ ) at $F = (\frac{25}{4}, 3)$ . Then $EF = \sqrt{(\frac{25}{4})^2 + (\frac{25}{3})^2} = 25\sqrt{\frac{1}{16} + \frac{1}{9}} = 25 \cdot \frac{5}{12} = \frac{125}{12}$ .

The trap

Assuming EF = 2 DB = 10 or that B is the midpoint of EF; BE and BF have different lengths and must be computed separately.

Common mistakes

Assuming EF = 2 DB = 10 or that B is the midpoint of EF; BE and BF have different lengths and must be computed separately.
Pairing the wrong sides in the similarity (scaling $AB$ by $\frac54$ and $CB$ by $\frac53$ ), which gives $5 + 5 = 10$ , choice (B).

Techniques

Set up the equation/formula and compute; no special trick needed