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Difference between revisions of "1959 AHSME Problems/Problem 40"

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== Solution ==
 
== Solution ==
  
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Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.
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Draw <math>\overline{DG} \parallel \overline{FC}</math> with <math>G</math> on <math>\overline{AB}</math>. We know that <math>GF = BF = 5</math>, since <math>\triangle BFE \sim \triangle BGD</math>.
  
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Since ADG is similar to ACB, we know that AG is 10.
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Likewise, since <math>\triangle ADG \sim \triangle ACF</math>, we know that <math>AG=5</math>.
  
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Thus, AF = 10 + 5 = 15, which is answer <math>\fbox{\textbf{(C)}}</math>.
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Thus, <math>AF=AG+GF+FB=5+5+5=15</math>, which is answer <math>\fbox{\textbf{(C)}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=39|num-a=41}}
 
{{AHSME 50p box|year=1959|num-b=39|num-a=41}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:33, 21 July 2024

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

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

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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All AHSME Problems and Solutions

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