2019 AMC 8 Problems/Problem 24

Problem 24

In triangle $ABC$ , point $D$ divides side $\overline{AC}$ s that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and left $F$ be the point of intersection of line $BC$ and line $AE$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$ ?

$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$A$",A,N); dot(B); label("$B$", B,SW);dot(C); label("$C$",C,SE); dot(DD); label("$D$",DD,NE); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]$

$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Solution 1

Draw $X$ so that $XD$ is parallel to $BC$ . That makes triangles $BEF$ and $EXD$ congruent since $BE$ = $ED$ . $FC$ =3 $XD$ so $BC$ =4 $BC$ . Since $AF$ =3 $EF$ ( $XE$ = $EF$ and $AX$ =1/3 $AF$ , so $XE$ = $EF$ =1/3 $AF$ ), the altitude of triangle $BEF$ is equal to 1/3 of the altitude of $ABC$ . The area of $ABC$ is 360, so the area of $BEF$ =1/3*1/4*360= $\boxed{(B) 30}$ ~heeeeeeeheeeee

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2019 AMC 8 Problems/Problem 24

Problem 24

Solution 1

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