1. The sum of the squares on the sides of a triangle is equal to twice the square on half the base, together with twice the square on the straight line which joins the vertex to the middle point of the base.
Let ABC be a triangle; and let D be the middle point of the base AB. Draw CE perpendicular to the base
meeting it at E; then B may be either in AB or in AB produced.
First, let B coincide with D; then the proposition follows immediately from I. 47.
Next, let E not coincide with D; then of the two angles ADC and BDC, one must be obtuse and one acute.
Suppose the angle ADC obtuse. Then, by II. 12, the square on AC is equal to the squares on AD, DC, together with twice the rectangle AD, DE; and, by II. 13, the square on BC together with twice the rectangle BD, DE is equal to the squares on BD, BC. Therefore, by Axiom 2, the squares on AC, BC, together with twice the rectangle BD, BE are equal to the squares on AD, DB, and twice the square on DC, together with twice the rectangle AD, DE. But AD is equal to DB. Therefore the squares on AC,BC are equal to twice the squares on AD, DC.
