1959 AHSME Problems/Problem 40

Problem

In $\triangle ABC$ , $BD$ is a median. $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$ . Point $F$ is on $AB$ . Then, if $\overline{BF}=5$ , $\overline{BA}$ equals: $\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}$

Solution

$[asy] import geometry; point A = (0,0); point B = (5,8); point C = (16,0); point D = midpoint(A--C); point E = midpoint(B--D); point F, G; triangle ABC = triangle(A,B,C); // Triangle ABC draw(ABC); dot(A); label("A",A,SW); dot(B); label("B",B,N); dot(C); label("C",C,SE); // Segment BD draw(B--D); dot(D); label("D",D,S); dot(E); label("E",E,SW); // Segment CF pair[] f = intersectionpoints(line(C,E),A--B); F = f[0]; dot(F); label("F",F,W); draw(C--F); // Segment DG pair[] g = intersectionpoints(parallel(D,line(F,C)),A--B); G = g[0]; dot(G); label("G",G,W); draw(D--G); // Length Label label("$5$", midpoint(B--F), NW); [/asy]$

Draw $\overline{DG} \parallel \overline{FC}$ with $G$ on $\overline{AB}$ . We know that $GF = BF = 5$ , since $\triangle BFE \sim \triangle BGD$ .

Likewise, since $\triangle ADG \sim \triangle ACF$ , we know that $AG=5$ .

Thus, $AF=AG+GF+FB=5+5+5=15$ , which is answer $\fbox{\textbf{(C)}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 41
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1959 AHSME Problems/Problem 40

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