is greater than the angle ACD, for a like reason the angle CBD is greater than the angle BCD; therefore the two angles CAD and CBD are together greater than the angle ACB, but they are also equal to it (I. 32, Cor. 4), which is impossible; therefore CD is not greater than AD or DB, and in the same manner it may be shewn that CD is not less than AD or DB; hence AD, DB, and DC are all equal.

EXERCISE XV.-THEOREM.

Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in that point.

Let ABC and DBE be two triangles having the same vertical angle DBC, and whose bases pass through the same point P, which bisects AC; to prove that the triangle BAC is less than the triangle BDE.

B

E

Through A draw AF parallel to BC; then in the triangles CEP and AFP, the angle ECP is equal to the angle FAP (I. 29), and the angles EPC, FPA are also equal (I. 15), also the side CP = AP, therefore the triangle EPC is equal to the triangle FPA, but the triangle ADP is greater than the triangle AFP, hence also the triangle ADP is greater than the triangle ECP, to each add the quadrilateral BAPE, and the triangle BDE is greater than the triangle BAC. In the same manner it can be proved that BAC is less than any other triangle whose base passes through P.

F

EXERCISE XVI.-THEOREM.

C

If two sides of a triangle be produced, the lines that bisect the two exterior angles, and the angle contained by the two sides produced, pass through the same point.

Let ABC be a triangle, having its sides AB and AC produced to P and Q; it is required to prove that the lines that bisect the three angles PBC, BCQ, and BAC, all pass through the same point.

Draw the lines BD and CD bisecting the angles PBC, BCQ, and intersecting in D; join AD, AD will bisect the angle BAC. Draw DG, DF, and DE perpendicular to

E

B

C

the sides; then in the triangles DGC, DFC, the angle DCG is equal to the angle DCF by construction, and the angles at G and Fare right angles, also the side DC is common and opposite to equal angles, therefore (I. 26) the triangles are equal in all respects; hence DGDF; in the same way it can be demonstrated from the triangles DEB and DFB that_DE= DF, and since DG and DE are each equal to DF, DG = DE. Again, in the two triangles DAG and DAE, right angled at G and E, the side DG has been proved equal to DE, and the hypotenuse DA is common, therefore they are equal in all respects (I. C), and the angle GAD is equal to EAD, hence AD bisects the angle BAC; and since there can only be one line drawn bisecting an angle, the lines BD, CD which bisect the two exterior angles, and the line AD which bisects the included angle, pass through the same point D.

EXERCISE XVII.-THEOREM.

If a right-angled triangle have one of the acute angles double of the other, prove that the hypotenuse is double of the side opposite the least angle.

Let ABD be an equilateral triangle, and draw AC bisecting the angle BAD; then BA and AC are equal to DA and AC, and the angle BAC = DAC, therefore BC = DC,

and the angle BCA is equal to the angle DCA, and they are adjacent angles, therefore each of them is a right angle. ABC is therefore a right-angled triangle, having the angle ABC double of the angle BAC, and it has been proved that BC, which is opposite the least angle, is half of BD, and B therefore half of the hypotenuse AB, which is equal to BD.

D

COR. If a right-angled triangle have one angle a third of a right angle, or 30°, the side opposite it is half of the hypotenuse.

EXERCISE XVIII.-THEOREM.

If the exterior angle, and one of the opposite interior angles, in one triangle, be respectively double those of another, the remaining opposite interior angle of the former is double that of the latter.

Let ABC and DEF be two triangles, of which the exterior angle CBG of the former is double the exterior angle FEH of the latter, and the interior angle A of the former double the interior angle D of the latter, then shall angle C of the former be double of F in the latter.

=

F

For angle CBG angles A and C together, also angle FEH = angles D and F together; but angle CBG= twice FEH by hypothesis; consequently angles A and C together must twice angle D and twice angle F; but A angle A is double of D by

A

B

G D

E

H

hypothesis; and taking away these equals from the preceding equal quantities, there remains angle C double of F.

Or more concisely thus:

but hence

angle CBG = A + C, angle FEH = D+F;

CBG twice FEH by hypothesis;

=

A+ C twice D and twice F;

but by hypothesis A

twice D; and taking away these equals from the preceding, there remains angle C = twice angle F.

EXERCISE XIX.-PROBLEM.

Through a given point to draw a line such that the segment of it intercepted between two given parallels may be equal to a given line.

Let P be the point, and AB, CD the parallels, and L the given line, to draw through P a line PAC, so that AC may be = L. Take any point B in one of the parallels, and from it as a centre, with L as a radius, cut CD in D; draw BD, and through P draw PC parallel to BD, and PC is the required line.

For AB, CD are given parallel, and PC is parallel to BD; hence the figure AD is a parallelogram, and consequently the side AC is = BD (I. 34).

EXERCISE XX.-PROBLEM.

P

L

A

B

D

Through a given point to draw a line that shall be equally inclined to two given lines.

Let AB, AC be two given lines, and P a given point, it is required to draw from P a line PC, making equal angles with AB and AC.

Produce CA to F, and bisect the angle BAF by the line AG, and from P draw PC parallel to AG, and it is the required line. For since AG and PC are parallel, therefore (I. 29) the exterior angle FAG is: = ACP,

the interior and opposite; and
GAB is = ABC, as they are
alternate angles; but FAG
GAB by construction, consequently F
ACP ABC.

When the point lies between the given lines, or is situated in one of them, it is evident that the same method of solution applies.

EXERCISE XXI.-THEOREM.

If two intersecting lines be respectively parallel, or equally inclined, to other two intersecting lines, the inclination of the former is equal to that of the latter.

CASE 1. When two of the intersecting lines are respectively parallel to the other two lines.

Let AC, AB be two intersecting lines respectively parallel to DF, DE; then angle A is = D.

For produce AB, DF, if necessary,

to meet in G; then the alternate angles
A and AGD are equal (I. 29), because AC
is parallel to DF; and the alternate Ꭺ.
angles D and AGD are also equal, since
AB is parallel to DE; consequently angles

AGD are equal.

F

G

B

D

E

CASE 2. Let the two intersecting lines AC, AG be respectively inclined to the two PE, PF, at the same angle; then angle AFPE.

For produce FP, EP, if necessary,

=

to meet the other two lines in C and
B; then, since the inclinations of FP,
EP, to AB, AC, respectively, are equal, A
therefore angle ACG GBP; but the
vertical angles AGC, PGB, at G, are also
equal; consequently, the third angles of
the triangles AGC, PGB, are equal; that
is, angle A = BPG (I. 32, Cor. 3) = FPE
(I. 15).

EXERCISE XXII.-THEOREM.

The sum of two sides of a triangle is greater than twice the line joining the vertex and the middle of the base.

Let ABC be a triangle, and CO the line joining the vertex and middle of the base; then AC and CB together are greater than twice CO.

For produce CO to D, till OD = OC, and join DB; then the sides AO, OC are equal to BO, OD, respectively, in the two triangles AOC, BOD, and the contained angles at O are equal; hence (I. 4) the triangles are every way equal, and there- A fore AC DB. But DB and BC together are greater than CD (I. 20); consequently, AC and BC are also greater than CD or twice CO.

EXERCISE XXIII.-PROBLEM.

Given the sum or difference of the hypotenuse and a side of a right-angled triangle, and also the remaining side, to construct it.

Let BC be a side of the right angle of a right-angled triangle, and AB the sum of the hypotenuse and the other side, to construct the triangle.

Let AB, BC be placed so that BC is perpendicular to AB; join AC, and bisect AC perpendicularly by DE; join EC, and EBC is the required triangle.

For in the triangles ADE, CDE, the sides AD, DE are respectively equal to CD, DE, and the angles contained by these sides are right angles; therefore

(I. 4) the triangles are every way equal, and hence CE = AE; therefore BE and EC together are equal to BA, the given sum, and BC is the given side; therefore CBE is the required triangle.

C

B

Again, let BC be the given side, and AB the given difference, to construct the triangle.

Let BC be placed perpendicularly to AB, and join AC; bisect AC perpendicularly by DE, and join CE; then EBC is the required triangle.

For, as in the preceding figure, the tri

=

angles ADE, CDE are every way equal, and therefore CE CA; consequently, the difference between CE and EB is equal to that between EA and EB; that is, it is AB. Hence AB is the difference between the E hypotenuse EC and the side EB, and BC is

the given side; therefore EBC is the required triangle.

C

B A

Otherwise, having placed the lines as before, and joined AC, make at the point C in the line AC the angle ACE = CAB, and produce the lines, if necessary, to meet in E; since the angles EAC and ECA are equal, the sides EA and EC are equal, and the remaining part of the proof is the same as above.

EXERCISE XXIV.-PROBLEM.

Through a given point, between two given lines, to draw a line so that the part of it intercepted between them may be bisected in that point.

Let AD, AC be the given lines, and P the given point.

Through P draw PB parallel to AD, make BC= AB; join CP, and produce CP to D, and it is the required line.