6. If a circle be described touching a semicircle and its diameter, the diameter of the circle is a harmonic mean between the segments into which the diameter of the semicircle is divided at the point of contact. 7. State and prove the Proposition corresponding to Ex. 5, for external contact of the circles. PROP. XIV.—Theorem. 1. Equiangular parallelograms (AB, CD) which are equal in area have the sides about the equal angles reciprocally proportional—AC : CE :: GC : CB. 2. Equiangular parallelograms which have the sides about the equal angles reciprocally proportional are equal in area. Dem.—Let AC, CE be so placed as to form one right line, and that the equal angles ACB, ECG may be vertically opposite. Now, since the angle ACB is equal to ECG, to each add BCE, and we have the sum of the angles ACB, BCE equal to the sum of the angles ECG, BCE; but the sum of ACB, BCE is [I. xiii.] two right angles. Therefore the sum of ECG, BCE is two right angles. Hence [I. xiv.] BC, CG form one right line. Complete the parallelogram BE. Again, since the parallelograms AB, CD are equal (hyp.), AB : CF :: CD : CF [V. vii.]; but AB : CF :: AC : CE [i.]; and CD : CF :: GC : CB [i.]; therefore AC : CE :: GC : CB; that is, the sides about the equal angles are reciprocally proportional. 2. Let AC : CE :: GC : CB, to prove the parallelograms AB, CD are equal. Dem.—Let the same construction be made, we have AB : CF :: AC : CE [i.]; and CD : CF :: GC : CB [i.]; but AC : CE :: GC : CB (hyp.). Therefore AB : CF :: CD : CF. Hence AB = CD [V. ix.]; that is, the parallelograms are equal. Or thus: Join HE, BE, HD, BD. The HC = twice the △ HBE, and the CD = twice the △ BDE. Therefore the △ HBE = BDE, and [I. xxxix.] HD is parallel to BE. Hence 2. May be proved by reversing this demonstration. Another demonstration of this Proposition may be got by producing the lines HA and DG to meet in I. Then [I. xliii.] the points I, C, F are collinear, and the Proposition is evident. PROP. XV.—Theorem. 1. Two triangles equal in area (ACB, DCE), which have one angle (C) in one equal to one angle (C) in the other, have the sides about these angles reciprocally proportional. 2. Two triangles, which have one angle in one equal to one angle in the other, and the sides about these angles reciprocally proportional, are equal in area. Dem.—1. Let the equal angles be so placed as to be vertically opposite, and that AC, CD may form one right line; then it may be demonstrated, as in the last Proposition, that BC, CE form one right line. Join BD. Now since the triangles ACB, DCE are equal, ACB : BCD :: DCE : BCD [V. vii.]; but ACB : BCD :: AC : CD [i.], and DCE : BCD :: EC : CB [i.]. Therefore AC : CD :: EC : CB; that is, the sides about the equal angles are reciprocally proportional. 2. If AC : CD :: EC : CB, to prove the triangle ACB equal to DCE.