12. True. Just divide the result from #11 by 2 to get semi-circle results.

15. The excircle proof is just the same as the incircle proof. Hint: the sides of an angle are equidistant from any point on the angle bisector. Now, write it up yourself.

19. a) Any point on the perpendicular bisector of a segment is equidistant from the endpoints. Therefore, the point at the intersection of two perpendicular bisectors of sides of a triangle must be equidistant from the three vertices. So this point of intersection is the center of a circle through the three vertices. b) Choose three points on the given arc and construct the circumcircle of those three points by constructing the perpendicular bisectors of two segments formed by the three points. c) 3 non-collinear points determine a circle. Construct the circumcircle of the three points and there it is.

because all the sides of a rhombus are equal. Calculate the midpoints of the diagonals to see they are the same. Then calculate the slopes of the diagonals. Their product is -1 when you cash in the Pyth Theorem: (d/(c-a)) * (d/(c+a)) = -1. Therefore the diagonals are perpendicular.

16. Place your right triangle in the first quadrant with the right angle at the origin. With vertices (0,0), (a,0) and (0,b), you can calculate the coordinates of the midpoint easily: (a/2,b/2). Now just use the distance formula on the segments from the midpoint to each vertex. The distance is always the sqrt(a^2 + b^2)/2.