Dem.—Erect CE at right angles to AB, and make it equal to AC or CB. Join AE, EB. Draw DF parallel to CE, and produce EB. Now since DF is parallel to EC, the angle BDF is = to BCE [I. xxix.], and [I. xv.] the angle DBF is = to EBC; but the sum of the angles BCE, EBC is less than two right angles [I. xvii.]; therefore the sum of the angles BDF, DBF is less than two right angles, and therefore [I., Axiom xii.] the lines EB, DF, if produced, will meet. Let them meet in F. Through F draw FG parallel to AB, and produce EC to meet it in G. Join AF. Because AC is equal to CE, and the angle ACE is right, the angle CEA is half a right angle. In like manner the angles CEB, CBE are half right angles; therefore the whole angle AEF is right. Again, because GF is parallel to CB, and GE intersects them, the angle EGF is equal to ECB [I. xxix.]; but ECB is right (const.); therefore EGF is right, and GEF has been proved to be half a right angle; therefore [I. xxxii.] GFE is half a right angle, and therefore [I. vi.] GE is equal to GF. In like manner FD is equal to DB. Again, since AC is equal to CE, AC2 is equal to CE2; but AE2 is equal to AC2 + CE2 [I. xlvii.]; therefore AE2 is equal to 2AC2. In like manner EF2 is equal to 2GF2 or 2CD2; therefore AE2 + EF2 is equal to 2AC2 + 2CD2; but AE2 + EF2 is equal to AF2 [I. xlvii.]. Therefore AF2 is equal to 2AC2 + 2CD2. Again, since DF is equal to DB, DF2 is equal to DB2: to each add AD2, and we get AD2 + DF2 equal to AD2 + DB2; but AD2 + DF2 is equal to AF2; therefore AF2 is equal to AD2 + DB2; and AF2 has been proved equal to 2AC2 + 2CD2. Therefore AD2 + DB2 is equal to 2AC2 + 2CD2. Or thus: AD = CD + AC, BD = CD − AC. Square and add, and we get AD2 + BD2 = 2CD2 + 2AC2. The following enunciations include Propositions ix. and x.:— 1. The square on the sum of any two lines plus the square on their difference equal twice the sum of their squares. 2. The sum of the squares on any two lines it equal to twice the square on half the sum plus twice the square on half the difference of the lines. 3. If a line be cut into two unequal parts, and also into two equal parts, the sum of the squares on the two unequal parts exceeds the sum of the squares on the two equal parts by the sum of the squares of the two differences between the equal and unequal parts. Exercises . 1. Given the sum or the difference of any two lines, and the sum of their squares; find the lines. 2. The sum of the squares on two sides AC, CB of a triangle is equal to twice the square on half the base AB, and twice the square on the median which bisects AB. 3. If the base of a triangle be given both in magnitude and position, and the sum of the squares on the sides in magnitude, the locus of the vertex is a circle. 4. If in the △ABC a point D in the base BC be such that prove that the middle point of AD is equally distant from B and C.