From A one of the angles of the triangle ABC let AD be drawn parallel to BC the opposite side; and from any point D in it, let DE, DH be drawn making any angles with the sides; draw BF, CG parallel to them respectively; DE: DH:: BF: CG.

E

H

Since DE is parallel to BF, and DA to BC, the triangles DEA, BFC are equiangular,

.. DE: DA :: BF ; BC;

and in a similar manner it may be shewn, that DA: DH:: BC: CG,

.. DE: DH :: BF: CG.

(15.) To bisect a given triangle by a line drawn from one of its angles.

Let ABC be the given triangle, and A the angle, from which the bisecting line is to be drawn. Bisect the opposite side AC in D, and join AD; AD bisects the triangle.

B

For the bases BD, DC being equal, (Eucl. i. 38.) the triangles ABD, ADC are also equal.

(16.) To bisect a given triangle by a line drawn from a given point in one of its sides.

Let ABC be the given triangle, and P the given point. Bisect BC in D, join AD, PD; and from A draw AE parallel to PD; join PE; PE bisects the triangle ABC.

P

Since AE is parallel to PD, the triangle APD is equal to the triangle EPD; from each of them take away the triangle PFD, and AFP EFD. Also since BD is equal to DC, the triangle ABD is equal to the triangle ADC; parts of which EFD, AFP are equal, .. ABEF is equal to PFDC; whence ABEF and AFP together, or ABEP will be equal to PFDC and FED together, i. e. to PEC; and .. the triangle ABC is bisected by PE.

(17.) To determine a point within a given triangle, from which lines drawn to the several angles, will divide. the triangle into three equal parts.

Let ABC be the given triangle; bisect AB, BC, in E, and D; join AD, CE, BF; Fis the point required.

=

B

E

Since BD DC, the triangle BAD is equal to DAC; and for the same reason the triangle BFD is equal to DFC; .. the triangle BFA is equal to AFC. Again, since BE=EA, the triangle BEC is equal to the triangle AEC; parts of which, the triangles BEF, AEF are equal; .. the triangle BFC is equal to AFC; and.. the three BFC, BFA, AFC are equal to one another.

(18.) To trisect a given triangle from a given point within it.

Let ABC be the given triangle, and P the given point within it. Trisect the side BC in D and E; join

PD, PE; and from 4 draw AF,
AG respectively parallel to them.
Join PF, PG, AP. Those three

e triangle into three

lines will divide equal parts. Join AD, AE. Since AF is parallel to PD, the triangle APF is equal to ADF; to each of these add ABF, .. APFB is equal to ADB. In the same manner APGC is equal to AEC; and .. the remainder FPG is equal to DAE. Now the triangles ABD, ADE, AEC, being on equal bases and of the same altitude, are equal, .. APFB, PFG, APGC are also equal; and the triangle ABC is trisected.

(19.) From a given point in the side of a triangle, to draw lines, which will divide the triangle into parts which shall have a given ratio.

Let ABC be the given triangle, and P the given point in the side BC. Divide BC, in the points D, E, F, into parts which shall have the given ratio. Join AD,

H

AE, AF, AP; and draw DG, EH, FI parallel to AP. Join PG, PH, PI; they will divide the triangle, as required.

For the triangles ABD, ADE, AEF, AFC being as their bases will be in the given ratio. And since DG is parallel to AP, the triangles DGA, DGP are equal, :. DBA, GPB are equal. And since the triangle ADP AGP, and AEP = AHP, .. ADE=HPG. Also APE=AHP, and APF=AIP, .. AEF=AHPI, and

=

AFC PIC; .. the parts PBG, GPH, HPIA, IPC are equal to ABD, ADE, AEF, AFC, and are.. in the given ratio. The same may be proved whatever be the number of parts.

(20.) If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines, a line be drawn to the opposite angle of the triangle; it will bisect that angle.

E

Let the exterior angles EBC, BCF, of the triangle ABC, be bisected by the lines BD, CD meeting in D. Join DA; it will bisect the angle BAC.

Let fall the perpendiculars DE, DF, DG. Then the angles DBE,

B

DBG being equal, and the angles at E and G being right angles, and DB common to the triangles DBE, DBG, .. DE=DG. In the same manner DG=DF; and .. DE=DF. Hence in the right-angled triangles DAE, DAF, DE is equal to DF and DA is common, .. the triangles are equiangular, and the angles DAE DAF are equal, i. e. BAC is bisected by AD.

(21.) If in two triangles the vertical angle of the one be equal to that of the other, and one other angle of the former be equal to the exterior angle at the base of the latter; the sides about the third angle of the former shall be proportional to those about the interior angle at the base of the latter.

Let ABC, DEF be two triangles having the angle BAC equal to EDF, and ABC equal to the exterior angle DFG, made by producing the side EF;

then AC CB :: DE : EF.

At the point D in the line FD, make the angle FDG equal to the angle EDF or BAC, and meeting EF pro

да

duced in G. Since the angle FDG is equal to the angle BAC, and DFG is equal to ABC, .. the triangles ABC, DFG are equiangular, and

AC: CB: DG: GF.

But since the angle GDE is bisected by DF, .. (Eucl. vi. 3.)

DG: GF:: DÉ EF,

:

.. AC: CB :: DE : EF.

(22.) In a given triangle to draw a line parallel to one of the sides, so that it may be a mean proportional between the segments of the base.

Let ABC be the given triangle; in the base of which take a point E, such that AE may be to EC in the duplicate ratio of AC: CB; draw ED parallel to BC; ED is the line required.

E C

Since ED is parallel to BC, AE: ED :: AC: CB. But AE EC in the duplicate ratio of AC: CB, and