HZ. Moreover, since we have drawn ZD parallel to HE, the line ED is to DA as HZ to ZA (Prop. II.); and we have shown that CE is to ED as BH to HZ. Therefore the undivided line AB has been cut similarly to the given divided line AC. . . . . Q. E. F.

PROPOSITION XI.-PROBLEM.

Πρότασις ια'.—Δύο δοθεισῶν εὐθειῶν, τρίτην ἀνάλογον προσευρεῖν.

To find a third proportional to two given right lines. Statement.-Let AB and AC be the two given straight lines. It is required to find a third proportional to AB and AC.

Construction.-Place AB and AC so as to make any angle BAC with each other. Produce AB and AC to the points D and E; making BD equal to AC (I. 3). Join BC, and through D draw DE parallel to BC (I. 31). The straight line CE is a third proportional to AB and AC.

D

B

G

Proof.—Because BC is parallel to DE, a side of the triangle ADE, AB is to BD as AC is to CE (Prop. II.). But BD is equal to AC. Therefore AB is to AC as AC is to CE.

Wherefore, to the two given straight lines AB and AC, a third proportional CE has been found. Q. E. F.

PROPOSITION XII-PROBLEM.

Πρότασις ιβ'.—Τριῶν δοθεισῶν εὐθειῶν, τετάρτην ἀνάλογον προσευρεῖν.

To find a fourth proportional to three given lines.

Statement.-Let F, E, and G be three given straight lines. It is required to find a fourth proportional to F, E, and G.

Construction.-Draw two lines AD and AI making

any angle; in AD take AB and BD equal to F and E, and in AI take AC equal to G; join

BC, and draw through D, DI parallel Dr
to BC; CI is the fourth proportional to
F, E, and G.

Proof. For in the triangle DAI, BC is parallel to DI, therefore AB is to BD as AC to CI (Prop. II.); but

B

FEG

the given lines F, E, and G are equal to AB, BD, and AC; therefore F is to E as G is to CI.

Therefore a fourth proportional has been found to the lines, F, E, G. . . . . Q. E. F.

PROPOSITION XIII.-PROBLEM.

Πρότασις ιγ ́.—Δύο δοθεισῶν εὐθειῶν, μέσην ἀνάλογον προσευρεῖν.

To find a mean proportional between two given straight lines.

Statement.-Let AB and BC be the two given straight lines. It is required to find a mean proportional between them.

Construction.-Place AB and BC in a straight line adjacent to each other; and upon AC describe the semicircle AIC. From the point B draw BI at right angles to AC (I. 11). The straight line BI is a mean proportional between AB and BC. Join AI and IC.

K

E.

F

Proof. Because the angle AIC in a semicircle is right (III. 31), and BI is drawn from the right angle perpendicular to the opposite side AC of the triangle AIC. Therefore IB is a mean proportional between AB and BC, the segments of the base (Prop. VIII., Cor.).

Wherefore, between the two given straight lines AB and BC a mean proportional IB has been found. Q. E. F.

PROPOSITION XIV.—THEOREM.

Πρότασις ιδ'. Τῶν ἴσων τε καὶ ἴσογωνίων παραλληλογράμ· μων, ἀντιπεπόνθασιν αἱ πλευραὶ, αἱ περὶ τὰς ἴσας γωνίας καὶ ὧν ἰσογωνίων παραλληλογράμμων, ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα.

Equal and equiangular parallelograms have their sides about the equal angles reciprocally proportional; and conversely, equiangular parallelograms which have their sides about the equal angles reciprocally proportional are equal to one another.

PART I.

Statement.-Let HB and BE be equal parallelograms which have their angles at B equal. The sides of the parallelograms HB and BE about the equal angles are reciprocally proportional.

Construction. If the sides AB and BC were so placed that they should make one right line, and that the equal angles be vertically opposite; since ABD and DBC are equal to two right an

H

D

F

B

E

gles (I. 13), and GBC is equal to ABD; therefore GBC and DBC are equal to two right angles, and therefore GB and DB form one right line (I. 14): complete the parallelogram DC.

Proof. Since the parallelograms AD and GC are equal, and DC is another parallelogram, AD is to DC (Prop. I.), as GC to DC (V. 7); but AD is to DC as AB to BC, and GC is to DC as GB to BD (Prop. I.); therefore AB is to BC as GB to BD (V. 11).

Wherefore the sides of the parallelograms, HB and BE about their equal angles, are reciprocally proportional.

G

Statement.

PART II.

Next, let the sides AB and AC of the

Join

triangle ABC be cut proportionally in the points D and E, that is, so that BD is to DA as CE to EA. DE, and it is parallel to BC.

Proof. The same construction being made, because BD is to DA as CE is to EA. But BD is to DA, as the triangle BDE is to the triangle ADE (Prop. I.); and CE is to EA, as the triangle CDE is to the triangle ADE. Therefore the triangle BDE is to the triangle ADE, as the triangle CDE is to the triangle ADE (V. 11). Wherefore the triangles BDE and CDE have the same ratio to the triangle ADE. Therefore the triangle BDE is equal to the triangle CDE (V.9); and they are on the same base DE. But equal triangles on the same base and on the same side of it are between the same parallels (I. 39). Therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION III. THEOREM.

Πρότασις γ ́.—Εὰν τριγώνου γωνία δίχα τμηθῇ, ἡ δὲ τέμν ουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευ ραῖς· καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν τοῦ τριγώνου γωνίαν.

If any angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio to one another which the adjacent sides of the triangle have; and conversely, if the segments of the base have the same ratio to one another which the adjacent sides of the triangle have, the straight line drawn from the vertex to the point of section bisects the vertical angle.

PART I.

Statement.-Let ABC be the triangle, and let the angle BAC be bisected by the straight line AD. The segment BD is to the segment DC as the side BA is to the side AC.

Construction.-Through the point

B

D

E

There

C draw CE parallel to DA (I. 31); and let BA produced meet CE in E. Proof. Because the straight line AC meets the parallels AD and EC, the angle ACE is equal to the alternate angle CAD (I. 29). But the angle CAD is equal to the angle BAD. fore the angle BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD and EC, the exterior angle BAD is equal to the interior and opposite angle AEC (I. 29). But the angle ACE has been proved equal to the angle BAD. Therefore also the angle ACE is equal to the angle AEC, and the side AE to the side AC (I. 6). Because AD is drawn parallel to EC, one of the sides of the triangle BCE; therefore BD is to DC as BA to AE (Prop. II.). But AE is equal to AC. Therefore, BD is to DC as BA is to AC (V. 7).

PART II.

Statement. Next, let the segment BD be to the segment DC as the side BA is to the side AC; and let AD be joined. The angle BAC is bisected by the straight line AD.

Proof. The same construction being made; because BD is to DC as BA is to AC; and BD is to DC as BA is to AE, because AD is parallel to EC (Prop. IL). Therefore BA is to AC as BA is to AE, and AC is equal to AE (V. 9); therefore the angle AEC is equal