Art. 166. PROPOSITION XLIX. (Euc. VI. 15.)

ENUNCIATION (i). Two triangles having equal areas and having one angle of the one equal to one angle of the other have their sides about the equal angles reciprocally proportional.

(ii) Two triangles which have one angle in the one equal to one angle in the other and the sides about the equal angles reciprocally proportional are equal in area.

These propositions may be deduced from Proposition 48. For the parallelograms ABCD, BEFG, BEHC are the doubles of the triangles ABC, BEG, BEC respectively; and it is merely necessary to repeat the proofs of Prop. 48, substituting for each parallelogram the triangle which is its half.

D

H

A

B

G

Fig. 129.

E

Art. 167. EXAMPLES.

73. Triangles which have one angle in the one supplementary to one angle in the other and their sides about the supplementary angles reciprocally proportional are equal in area.

74. Triangles having equal areas and having one angle of the one supplementary to one angle of the other, have their sides about the supplementary angles reciprocally proportional.

75. If P be any point on the side AC of the triangle ABC, and if PQ be drawn parallel to BC to cut AB at Q, then if a straight line through P cut BA produced through A at R and BC at S so as to make the triangles ABC, BRS equal, prove that QR will be a third proportional to QA and QB.

76. The triangles ABC, DEF are similar, and on DE the side corresponding to AB a point K is taken so that DK is a third proportional to DE and AB, prove that the triangles ABC, DKF are equal in area.

(This is the proposition on which Euclid's proof that similar triangles are to one another in the duplicate ratio of corresponding sides is based.)

77. If a straight line DE be drawn parallel to the base BC of the triangle ABC cutting AB at D and AC at E, and if AF be drawn perpendicular to DE, prove that the rectangle AF, BC is double of the triangle AEB.

SECTION IX.

MISCELLANEOUS GEOMETRICAL PROPOSITIONS. Props. 50-55.

Art. 168. PROPOSITION L.

ENUNCIATION. If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base together with the square on the straight line which bisects the angle.

Let ABC be the triangle.

Bisect BAC by AD cutting the base BC at D.

It is required to prove that

rect. AB. AC=rect. BD. DC + square on AD. Describe a circle round the triangle ABC, and

let AD cut the circle at E.

Join CE.

In the triangles ABD, AEC

BAD = EAC

ABD = AEC, since they stand on the same arc AC. .. ADB= AĈE.

ENUNCIATION. If from any vertex of a triangle a perpendicular be drawn to the opposite side, the diameter of the circle circumscribing the triangle is a fourth proportional to the perpendicular and the sides of the triangle which meet at that

Hence by Prop. 26 the triangles are similar.

.. DB: CB = BA : BE AD: EC.

From the equality of the first and second ratios it follows that the diameter BE is a fourth proportional to the perpendicular BD and the sides BC, BA.

If D is any point on the side BC of a triangle ABC, then the diameters of the circles circumscribing the triangles ABD and ACD are proportional to the sides AB, AC.

Art. 171. PROPOSITION LII. (Euc. VI. 30.)

ENUNCIATION. To divide internally or externally a finite straight line in extreme and mean ratio; i.e. so that the whole line is to one segment as that segment is to the other segment.

Let AB be the straight line, it is required to find a point C on it so that AB: AC AC: CB,

On AG' describe the square AC'H'G'.

The points C, C′ fall on AB, and are the required points.
It is proved in Prop. 11 of the Second Book of Euclid that
the square on AC=rect. AB.BC.

.. AB: AC AC: BC.

To prove the same property for the point C'.
Since AE is bisected at F and produced to G',

[Cor. to Prop. 38.

.. square on FG'square on FA + rect. EG'. AG'.
.. square on FB square on FA+rect. EG'H'K'.

=

If ABC be a triangle right-angled at A, and AD be drawn perpendicular to the hypotenuse cutting it at D, and if D divide BC in extreme and mean ratio, then prove that the sides of the triangle ABC are in proportion.

Art. 173. PROPOSITION LIII. (Euc. VI. 24.)

ENUNCIATION. Parallelograms about the diagonal of any parallelogram are similar to the whole and to one another.

Let ABCD be a parallelogram.

Let AEFH, FKCG be parallelograms about the diagonal AC of the parallelogram ABCD.

It is required to prove that they are similar to ABCD and to one another.

Since EF is parallel to BC,

the triangles AEF, ABC are similar.

[Cor. to Prop. 26.

.. AE: AB = EF : BC = FA : CA. Since FH is parallel to CD, the triangles AFH, ACD are similar.

Hence

Further

A

E

H

F

[Cor. to Prop. 26.

.. FACA AH: AD = HF : DC.
AE AB EF: BC= FH: CD
: =

B

K

HA: DA.

[Prop. 10.

HÂE=DÂB,

AEF = ABC,

EFH = BĈD,

FĤA = CÔA.