remaining sides be also equal to one another, the figure is a parallelogram.

Let any two opposite sides, as AB, DC, of

D

B

the quadrilateral figure ABCD, be equal to one another, and let the two remaining sides, AD, BC, be, also, equal to one another: The figure ABCD is a .

For, join D, B: Then since the two sides AD, DB, of the ▲ ADB, are equal to the two sides CB, BD, of the ▲ CBD, and that the base AB is equal (hyp.) to the base DC, .*. (E. 8. 1.) the ▲ ADB=2 DBC; and (E. 4. 1.) the ▲ ABD= < BDC; .(E. 27. 1.) AD is parallel to BC, and AB is parallel to DC; i. e. the figure ABCD is a .

27. COR. 1. Hence may be deduced a practical method of drawing a straight line, through a given point, parallel to a given straight line.

For, let it be required to draw through the given point B, a straight line parallel to AD: From any point A in AD, as a centre, and at any distance, describe a circle cutting AD in D; and from B as a centre, at the same distance, describe another

circle; lastly, from D as a centre, at a distance equal to that of A, B, describe another circle, cutting the circle last described in C; join B, C. BC is parallel to AD.

For, if A, B and D, C be joined, it is manifest from the construction, that AD = BC, and AB=DC: ... (S. 16. 1.) BC is parallel to AD. 28. COR. 2. A rhombus is a parallelogram.

PROP. XIX.

29. THEOREM. Every parallelogram which has one angle a right angle, has all its angles right angles.

Let one ▲, as A, of the ABCD be a right angle: The B, C, and D are also right angles.

Al

B

For, since AD is parallel to BC, and AB meets them, the two interior A, B are, (E. 29. 1.) together, equal to two right; but (hyp.) the ▲ A is a right ; .. the B is also a right : And, in the same manner, may the remaining 4, C and D, be shewn to be right .

PROP. XX.

30. PROBLEM. To trisect a right angle; i. e. to divide it into three equal parts.

Let the XAY be a right : It is required

to trisect it; i. e. to divide it into three equal parts.

In AX take any point B; upon AB describe (E. 1. 1.) the equilateral A ACB; and from A draw (E. 12. 1.) AD 1 to BC: The XAY is trisected by the two straight lines AC and AD.

For, from C draw (E. 12. 1.) draw CE to AY; then, since the BAE, AEC, are right .*. (E. 28. 1.) AB is parallel to EC; ... (E. 29. 1.) LECA CAB ACB; because (constr.) the AACB is equilateral, and (E. 5. 1. cor.) equiangular: Since, ..., the ACE =▲ ACD,

and that the

is common to

(E. 26. 1.) the

D and E are right, and AC the two ADC, AEC,

EAC-DAC: Again, since

(constr. and E. 5. 1. cor.) the ACB=▲ ABC, and (constr.) the

at D are right angles, and that AC=AB, ... (E. 26.1.) the DAC=2 DAB: But it was shewn that the EAC✩ DAC: .. EAC DAC= 2 DAB; i. e. the right <XAY is trisected by AC and AD.

PROP. XXI.

31. PROBLEM. Hence, to trisect a given rectilineal angle, which is the half, or the quarter, or the eighth part, and so on, of a right angle.

First, let the given YAZ, be the half of a

right, and let it be required to trisect it.

Draw (E. 11. 1.) from A, AX AY; trisect (S. 18.1.) the right XAY; then (S. 1. 1.) trisect the YAZ, which is the half of the YAX.

But, if the given be the quarter of a right angle, its double may be trisected by the former case; and... the given itself may be trisected

by (S. 1. 1.)

And, by following the same method, it is evident that an may be trisected, which is the eighth part, or the sixteenth part, and so on, of a right angle.

PROP. XXII.

32. PROBLEM. In the hypotenuse of a right-angled triangle, to find a point, the perpendicular distance of which from one of the sides, shall be equal to the segment of the hypotenuse between the point and the other side.

Let ABC be a right-angled A, right-angled

at C: It is required to find a point in the hypo