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Difference between revisions of "1980 AHSME Problems/Problem 21"

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== Solution 1 ==  
 
== Solution 1 ==  
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Let <math>M</math> be the midpoint of <math>\overline{DC}</math>.  Then <math>\triangle ECM ~ \triangle ACD</math>,  
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Let <math>M</math> be the midpoint of <math>\overline{DC}</math>.  Then <math>\triangle ECM \sim \triangle ACD</math> and <math>\overline{EM} || \overline{AD}</math>.  Since <math>\overline{EM} || \overline{FD}</math>, it follows that <math>\triangle BFD \sim \triangle BEM</math>.  Let <math>a</math> be the area of <math>\triangle BFD</math>.  Since the sides of <math>\triangle BEM</math> are twice as long as the corresponding sides of <math>\triangle BFD</math>, the area of <math>\triangle BEM</math> must be <math>2^2=4</math> times the area of <math>\triangle BFD</math>, that is, <math>4a</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 03:55, 25 June 2025

Problem

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy]

In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$, and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$. The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is

$\text{(A)} \ \frac 15 \qquad \text{(B)} \ \frac 14 \qquad \text{(C)} \ \frac 13 \qquad \text{(D)}\ \frac{2}{5}\qquad \text{(E)}\ \text{none of these}$


Solution 1

Let $M$ be the midpoint of $\overline{DC}$. Then $\triangle ECM \sim \triangle ACD$ and $\overline{EM} || \overline{AD}$. Since $\overline{EM} || \overline{FD}$, it follows that $\triangle BFD \sim \triangle BEM$. Let $a$ be the area of $\triangle BFD$. Since the sides of $\triangle BEM$ are twice as long as the corresponding sides of $\triangle BFD$, the area of $\triangle BEM$ must be $2^2=4$ times the area of $\triangle BFD$, that is, $4a$.

Solution 2

We can use the principle of same height same area to solve this problem. $\fbox{A}$

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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