# Difference between revisions of "1978 AHSME Problems/Problem 23"

## Problem

[asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("$A$",(0,0),SW); label("$B$",(1,0),SE); label("$C$",(1,1),NE); label("$D$",(0,1),NW); label("$E$",(.5,sqrt(3)/2),E); label("$F$",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy] Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$, and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$. If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$

## Solution

No solutions yet!