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