<Dd> 7a . Construct a vertical line through F. It intersects the original circle at two of the vertices of the pentagon . The third vertex is the rightmost intersection of the horizontal line with the original circle . </Dd> <Dl> <Dd> 8a . Construct the other two vertices using the compass and the length of the vertex found in step 7a . </Dd> </Dl> <Dd> 8a . Construct the other two vertices using the compass and the length of the vertex found in step 7a . </Dd> <Ol> <Li> We first note that a regular pentagon can be divided into 10 congruent triangles as shown in the Observation . Also, cos 36 ° = 1 + 5 4 (\ displaystyle (\ tfrac (1 + (\ sqrt (5))) (4))). † </Li> <Li> In Step 1, we use four units (shown in blue) and a right angle to construct a segment of length 1 + √ 5, specifically by creating a 1 - 2 - √ 5 right triangle and then extending the hypotenuse of √ 5 by a length of 1 . We then bisect that segment--and then bisect again--to create a segment of length 1 + 5 4 (\ displaystyle (\ tfrac (1 + (\ sqrt (5))) (4))) (shown in red .) </Li> <Li> In Step 2, we construct two concentric circles centered at O with radii of length 1 and length 1 + 5 4 (\ displaystyle (\ tfrac (1 + (\ sqrt (5))) (4))). We then place P arbitrarily on the smaller circle, as shown . Constructing a line perpendicular to OP passing through P, we construct the first side of the pentagon by using the points created at the intersection of the tangent and the unit circle . Copying that length four times along the outer edge of the unit circles gives us our regular pentagon . </Li> </Ol>

Are all the sides of a pentagon equal