Then no other straight line, terminated by the sides AB, AC, can be bisected in the point F.

Proof.-Draw any other straight line GFH, and, if possible, suppose it bisected in the point F.

In the triangles DFG, EFH, DF is equal to EF (hyp.), and GF is equal to FH;

therefore DF, FG are equal to EF, FH, each to each; and the angle DFG is equal to the angle EFH (1. 15); therefore the base DG is equal to the base EH (1. 4), and the two triangles are equal in every respect; wherefore the angle DGF is equal to the angle EHF. Because GH meets the two straight lines AB, AC, and makes the alternate angles BGH, AHG equal to one another, therefore AB, AC are parallel (1. 27);

but they are not parallel (hyp.);

therefore GH is not bisected in F.

Wherefore DE is the only straight line terminated by AB, AC, which is bisected in F. Q.E.D.

A straight line is drawn bisecting an angle of a parallelogram: prove that there are formed by it and the sides of the 'parallelogram (or those produced) three triangles which are isosceles.

Let ABCD be a parallelogram, and let the angle ABC be bisected by the straight line BE, cutting AD in F, and meeting CD produced in E.

Then the three triangles ABF, EBC, EFD shall be isosceles.

Proof. Because BF meets the parallel straight lines AD, BC,

the alternate angles AFB, FBC are equal (1. 29);

but the angle ABF is equal to the angle FBC (hyp.);

therefore the angle AFB is equal to the angle ABF (ax. 1), and therefore the side AF is equal to the side AB (1. 6). Wherefore the triangle ABF is isosceles.

Again, because BE meets the parallel straight lines CE, AB, the alternate angles BEC, ABE are equal;

but the angle EBC is equal to the angle ABE;

therefore the angle BEC is equal to the angle EBC,

and therefore the side CE is equal to the side BE. Wherefore the triangle EBC is isosceles.

Again, because BE falls upon the parallel straight lines AD, BC,

the exterior angle EFD is equal to the interior and opposite

angle EBC (I. 29);

but it has been proved that the angle FED is equal to the angle EBC;

therefore the angle EFD is equal to the angle FED,

and therefore the side DF is equal to the side DE. Wherefore the triangle EFD is isosceles.

Therefore the three triangles ABF, EBC, EFD are isosceles. Q.E.D.

29. In the triangle ABC, AD bisecting the angle BAC

meets BC in D, and DE, DF parallel to AC, AB respectively, meet AB, AC in E, F. Show that DE, DF are equal.

Proof. Because AD meets the parallel straight lines AC, DE,

the alternate angles DAF, ADE are equal (1. 29);

but the angle DAE is equal to the angle DAF (hyp.); therefore the angle EDA is equal to the angle DAE (ax. 1); and therefore the triangle EAD is isosceles (1. 6).

Similarly it can be proved that the triangle FAD is isosceles. Now, in the triangles EAD, FAD,

the angles EAD, EDA are equal to the angles FAD, FDA, each to each,

and the side AD is common to the two triangles;

wherefore the side DE is equal to the side DF (1. 26). Q.E.D.

AB is parallel to CD, AD is bisected in E: show that any other straight line drawn through E to meet the two lines will be bisected in that point.

Proof-Draw any other straight line through E, as FEG. Because AB is parallel to CD, and FG meets them, the angle AFE is equal to the alternate angle EGD (1. 29);

and the angle AEF is equal to the angle DEG (1. 15). Hence in the triangles AEF, DEG,

the two angles AFE, AEF are equal to the two angles DGE, DEG, each to each,

and the side AE is equal to the side DE (hyp.);

therefore the side EF is equal to the side EG (1. 26):

wherefore GE is bisected in F.

Therefore every straight line drawn through E to meet AB, CD, is bisected in E. Q.E.D.

31. From a given point draw a straight line such that the part of it included between two given parallel straight lines shall be of a given length. In what case would the construction fail?

Let AB, CD be the two given parallel straight lines, E the given point, and F the given length.

It is required to draw a straight line from the point E such that the part of it included between the parallel straight lines AB, CD shall be equal to F.

In AB take any point G, and make GO equal to F (1. 3). From the centre G, at the distance GO, describe a circle cutting CD in H (post. 3), and join GH (post. 1).

Through E draw EK parallel to GH (1. 31), and cutting AB in L.

Then LK shall be equal to F.

Proof-Because LKHG is a parallelogram (con.), therefore LK is equal to GH (I. 34);

but GH is equal to GO (def. 15), and therefore equal to F (ax. 1); therefore LK is equal to F.

Wherefore from the given point E a straight line EK has been drawn, such that the part LK included between the given parallel straight lines AB, CD is equal to the given length F. Q.E.F

Note.—The construction would fail if the given length were less than the perpendicular distance between AB, CD.

Draw a line DE parallel to the base BC of a triangle ABC, so that DE is equal to the difference of BD and CE.

E

B

Let AB be the greater of the two sides.

From AB cut off AF equal to AC (1. 3), and join CF (post. 1); bisect the angle ABC by the straight line BG (1. 9), cutting

CF in G;

through G draw GE parallel to AB (1. 31), cutting AC in E, and through E, G draw ED, GH parallel to BC.

Then DE shall be equal to the difference of BD, CE.

Proof. Because AF is equal to AC (con.),

therefore the angle AFC is equal to the angle ACF (1. 5); but the angle EGC is equal to the angle AFC (1. 29); therefore the angle EGC is equal to the angle ECG (ax. 1); wherefore EG is equal to EC (1. 6);

but DH is equal to EG (1. 34);

therefore DH is equal to EC:

wherefore BH is the difference between BD, CE.

Again, the angle HGB is equal to the angle GBC (1. 29);