But HD is the rectangle contained by AB, BH, for AB is
equal to BD ;
and FH is the square on AH;
therefore the rectangle AB,BH is equal to the square on AH.
Wherefore the straight line AB is divided at H, so that the rectangle AB, BH is equal to the square on AH. q.e.f.
PROPOSITION 12. THEOREM.
In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, then produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle.
Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpendicular to BC produced ; the square on AB shall be greater than the squares on AC, CB, by twice the rectangle BC, CD.
Because the straight line
BD is divided into two parts
it the point C, the square on
BD is equal to the squares on
BC, CD, and twice the rectangle
BC, CD. [II. 4.
To each of these equals add the
quare on DA.
Therefore the squares on BD, DA are equal to the squares on
BC, CD, DA, and twice the rectangle BC, CD. [Axiom 2.
But the square on BA is equal to the squares on BD, DA,
because the angle at D is a right angle ; [I. 47.
and the square on CA is equal to the squares on CD, DA. [1. 47.
therefore the square on BA is equal to the squares on
BC, CA, and twice the rectangle BC, CD ;
that is, the square on BA is greater than the squares on
BC, CA by twice the rectangle BC, CD.
Wherefore, in obtuse-angled triangles &c. q.e.d.
