PROPOSITION III.

If the line bisecting an exterior angle adjacent to the vertical angle of a triangle meet the base produced, then will the segments between the dividing line and the extremities of the base have the same ratio which the sides of the triangle have to one another.

Let AQ bisect the exterior CAX of ▲ ABC and cut BC produced in Q.

Then BQ QC as BA: AC.

Through C draw CR || to QA.

Now CR is to QA,..

XAQ = 4 ARC, ((1. 21)

and

CAQ = LACR; (1.21)

but 4 XAQ= CAQ, .. LARC = LACR;
4

Converse Theorem.

Let Q be a point, in the side BC of ▲ ABC produced, such that BQ QC as BA AC. Then shall AQ bisect the exterior L CAX adjacent to L BAC.

.. LACR is = L CAQ and ▲ ARC is = XAQ;

.. ▲ CAQ=XAQ.

(I. 21)

PROPOSITION IV.

The sides about the equal angles of equiangular triangles are proportional.

F

B

H

Let ABC, FGH be equiangular triangles. Then shall the sides about the equals A and F be proportional,

as also the sides about B and G, and those about C and H.

From the arms of the 4 at A, produced if necessary,
cut off AQ=FG and AK = FH; join QK.

Then As AQK, FGH are equal in all respects; (I. 1)
.. LAQK = L FGH, and ... = LABC;

.. BC is to QK;

.. AB: AQ as AC : AK;
.. AB: AC as AQ : AK,
and .. as FG: FH.

(I. 20)

(VI. I)

(v. 6)

Similarly it may be shewn that the sides about B and G are proportional, as also those about C and H.

Conversely,

If the sides about the angles of one triangle be proportional to the sides about the angles of another, the triangles shall be equiangular.

A

F

H

Let ABC, FGH be two As, such that

AB: AC as FG : FH,

and AB BC as FG: GH.

The As ABC, FGH shall be equiangular.

From the arms of the at A, produced if necessary, cut off AQ=FG, and AK = FH; join QK.

.. ▲ FGH is equal to ▲ AQK in all respects;

Also,

AFGH is equiangular to ▲ ABC.

(v. 4)

(1.5)

If two As have one angle of the one equal to one angle of the other, and the sides about the equal angles proportional, then shall the As be equiangular.

Let ABC, FGH be two As, having 4 BAC = L GFH,

and such that AB AC as FG: FH.

Then shall as ABC,

FGH be equiangular.

From the arms of the

at A, produced if necessary,

cut off AQ=FG and AK = FH; join QK.

Then As AQK, FGH are equal in all respects,

and. AB: AC as FG: FH,

PROPOSITION V.

If through any point within or without a circle two secants be drawn, their segments between the given point and the circumference shall be reciprocally proportional.

R, S.

PQR,

let two secants be drawn cutting the Oce in P, Q and

Then XP: XS as XR : XQ.

Join PS, RQ.

Then in As PXS, RXQ,

L PXS= L RXQ, and ▲ PSX = ▲ RQX;

L

.. ▲s PXS, RXQ are equiangular;

.. XP: XS as XR: XQ.

(III. 14)

(I. 25)

(VI. 4)

Rectangle XP, XQ=rectangle XR, XS. (v. 5)

COR. 1.

COR. 2.

Let the point X be without

the O, and let XPQ revolve about X until

P and Q coincide,

then the secant XPQ becomes a tangent, and rectangle XP, XQ becomes the square on XQ;

.. rectangle XR, XS=square on XQ.

P