1959 AHSME Problems/Problem 40

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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 1

[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 $\fbox{\textbf{(C)}}$.


Solution 2

Let $AD=DC=x$ and $BE=ED=y$. By Menelaus' Theorem on $\triangle BEF$ and $\overleftrightarrow{AC}$, we know the following: BDEDECFCFABA=12yyECFCFA5+FA=1ECFCFA5+FA=12 By applying Menelaus again on $\triangle CDE$ and $\overleftrightarrow{AB}$, we see that: CADADBEBEFCF=12xx2yyCFECFC=11ECFC=14ECFC=34 Substituting $\frac{3}{4}$ for $\frac{EC}{FC}$ into the previous equation, we can now solve for $FA$: ECFCFA5+FA=1234FA5+FA=12FA5+FA=233FA=10+2FAFA=10 Because $AB=AF+FB$, $AB=5+10=\boxed{\textbf{(C) } 15}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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