1990 AHSME Problems/Problem 20

Problem

$[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.1; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]$

In the figure $ABCD$ is a quadrilateral with right angles at $A$ and $C$ . Points $E$ and $F$ are on $\overline{AC}$ , and $\overline{DE}$ and $\overline{BF}$ are perpendicual to $\overline{AC}$ . If $AE=3, DE=5,$ and $CE=7$ , then $BF=$

$\text{(A) } 3.6\quad \text{(B) } 4\quad \text{(C) } 4.2\quad \text{(D) } 4.5\quad \text{(E) } 5$

Solution

Label the angles as shown in the diagram. Since $\angle DEC$ forms a linear pair with $\angle DEA$ , $\angle DEA$ is a right angle.

$[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.075; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, A)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, C)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]$

Let $\angle DAE = \alpha$ and $\angle ADE = \beta$ .

Since $\alpha + \beta = 90^\circ$ , and $\alpha + \angle BAF = 90^\circ$ , then $\beta = \angle BAF$ . By the same logic, $\angle ABF = \alpha$ .

As a result, $\triangle AED \sim \triangle BFA$ . By the same logic, $\triangle CFB \sim \triangle DEC$ .

Then, $\frac{BF}{AF} = \frac{3}{5}$ , and $\frac{CF}{BF} = \frac{5}{7}$ .

Then, $7CF = 5BF$ , and $5BF = 3AF$ .

By the transitive property, $7CF = 3AF$ . $AC = AF + CF = 10$ , and plugging in, we get $CF = 3$ .

Finally, plugging in to $\frac{CF}{BF} = \frac{5}{7}$ , we get $BF = 4.2$ $\fbox{C}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Page

Toolbox

Search

1990 AHSME Problems/Problem 20

Problem

Solution

See also