<p>Can someone please explain number 20 of pg. 756 in the blue book? It's a picture one.</p>
<p>You probably mean q.20 p.796.</p>
<p>Oh yes, sorry.</p>
<p>all 5 line segments are equal in length (congruent), so abd and dbc are equilateral triangles. draw the segment AC, which will be perpendicular bisector to BD at a point we will call E. in this example we will work with triangle BEC. BC is 1, BE is 1/2. By pythagorean theorem EC is (sqrt3 / 2). AE is same length as EC. so double the length of EC is the length of AC. </p>
<p>AC is sqrt 3 : BD is 1</p>
<p>that probably wasnt very clear, sorry</p>
<p>No (it wasn't), that was very clear! Thanks a lot.</p>
<p>Based on the superlyduper's post:
AC/BD = CE/BE.
CEB is 30-60-90 triangle.
From the Special Right Triangles diagram at the beginning of each math section
CE/BE = x sqrt(3) / x = sqrt(3) / 1.
Answer B.</p>