and CE be drawn to AB and AC, and equal to AB and AÇ, respectively, and let AK be drawn 1 to BC: Then if D, C and E, B be joined, DC and EB shall cut one another in the same point of AK.

For, if it be possible, let DC cut AK in P, and let EB cut AK in H; and from D and E draw (E. 12. 1.) DF and EG 1 to BC produced both ways; .'. (S. 38. 1.) FB=GC, and ... FC=BG: And, since (constr.) the PKC, DFC, are right

, and that the PCK is common to the two APCK, DCF, .. (S. 26. 1.) the two PCK, DCF are equiangular; and, in the same manner, the two HKB, EGB may be shewn to be equiangular; .. (E. 4. 6.) CF: FD:: CK: KP.

But (S. 38. 1. cor.) FD=BK, and CK=GE; and it has been shewn that CF-BG;

..

But (E. 4. 6.)

BG; BK::GE: KP:

BG: BK::GE: KH;

.. (E. 9. 5.) KH=KP; which is absurd; .. DC and EB cannot cut the perpendicular drawn from A to BC, in two different points.

PROP. XIV.

20. THEOREM. The semi-circumference of a circle having been divided into any number of equal parts, and chords having been drawn, from either

extremity of the diameter, to the several points of division, the first chord has to the second, the same ratio which the second has to the aggregate of the first and third; or the same ratio which any other chord has to the aggregate of the two chords that are next to it.

Let the semi-circumference AEL of a circle,

be divided into any number of equal parts, in the points B, C, D, E, F, &c.; and let AB, AC, AD, AE, AF, &c., be drawn: Then

AB: AC:: AC: AB+ AD :: AD: AC+ AE, and

so on.

For, from C, as a centre, at the distance CA, describe a circle cutting AD, produced, in M, and join B, C, and C, D, and C, M; and since (hyp.) AB = BC, ... (E. 29. 3.) AB= BC; also (E. 27. 3.) the BAC=1 CAD; .. (E. 5. 1. and S. 26. 1.) the isosceles ABC, ACM, are equiangular;

.. (E. 4. 6.) AB: AC:: AC: AM, or DM+AD: But (E. 22. 3.) since ABCD is a quadrilateral figure inscribed in a circle, the ▲ ABC+ 2 ADC= two right; also (E. 13. 1.) the ADC+ ‹ two right ; .. the CDM=2 ABC; BAC CAD, or (constr. and E. 5. 1.)

CDM and the

2 CMD; and the side CM, of the ACDM, is equal to the side CA, of the ▲ ABC; .. (E. 26. 1.) DM=AB; and it has been shewn that AB: AC:: AC: DM+AD;

.. AB: AC:: AC: AB+AD:

And, by a similar construction, and a similar method of proof, the remaining part of the proposition may be demonstrated.

PROP. XV.

21. PROBLEM. From a given point, either within or without a given rectilineal angle, to draw a straight line cutting off from the lines which contain the angle, segments, towards the summit of the angle, which shall be to one another in a given

ratio.

Let PAQ be the given 2, and first let B be a given point without it: It is required to draw, from B, a straight line which shall cut off from AP and AQ, two segments, towards A, which shall be to one another in a given ratio.

From AP and AQ cut off AC and AD, equal

to the two straight lines which exhibit the given ratio, each to each; join D, C; and through B draw (E. 31.1.) BEF parallel to DC: Then, since (constr. and E. 29. 1.) the two ▲ ADC, AEF are equiangular,

.`. (E. 4. 6.) AF:AE::AC: AD;

.. (constr.) AF is to AE in the given ratio. And the problem may be solved in the same manner, when the given point is within the given angle.

PROP. XVI.

22. PROBLEM. To draw through a given point a straight line cutting the lines which contain a given rectilineal angle, so that the segment of it, between those lines, shall be divided by the straight line that bisects the given angle, into two parts, which are to one another in a given ratio.

Let PAQbe the given 4; let AR be drawn (E. 9.

1.) bisecting it; and let B be the given point: It is required to draw, from B, a straight line cutting AP and AQ, so that the segment of it, between AP and AQ, shall be divided by AR, into two parts, which are to one another in a given ratio.

Draw (S. 15. 6.) BEF, so that AF shall be to AE in the given ratio, and let BF cut AR in H; then, (E. 3. 6.) since AH bisects the FAE, FH :HE:: AF: AE; that is, FG is to GE in the given ratio.

PROP. XVII.

23. THEOREM. If two trapeziums have an angle of the one equal to an angle of the other, and if, also, the sides of the two figures, about each of their angles, be proportionals, the remaining angles of the one shall be equal to the remaining angles of the other.