The figure on the left shows a geometrical demonstration of Euclid's second theorem (Corollary to Proposition 8 of Book VI): " If in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the straight line so drawn is a mean proportional between the segments of the base "
	Let the triangle be ABC, with right angle C. AB is the base and CH is the height. The proposition says that AH : CH = CH : HB. This means that the square over CH has the same area as the rectangle with sides AH and HB. By rotating the triangle HAC 90° clockwise around H we get the triangle HA"C", and rotating it counterclockwise we get HA'C'. Thus C'H and HC are two sides of the square C'HCE. Since the triangles HA"C" and HBC are similar, A"C" and BC are parallel. Let A"' and B" be the intersections of the line through A" parallel to the base AB, with the lines EC' and CB, respectively. Since A'H=A"H, the rectangle HA'B'B has the same area as HBB"A"'. This parallelogram, HBB"A"', can be divided into two pieces. A"'HC"A" has the same area as A"'C'HA". A"C"BB" has the same area as A"C"DC. Since C'H is equal to HC", A"C"DC has the same area as A"'A"CE. Did you like it ? Try some other trivial math quiz ... Or read about math topics.