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THEOR. 4. If two triangles have the sides taken in order about each of their angles proportional, they are similar, and those angles which are opposite to the homologous sides are equal.
Let ABC, DEF be two triangles having the side AB to the side BC as the side DE to the side EF, the side BC to the side CA as the side EF to the side FD, and consequently, ex aquali, the side AB to the side AC as the side DE to the side DF:
then shall the angle ABC be equal to the angle DEF, the angle BCA to the angle EFD, and the angle CAB to the angle FDE.
In ED make EH equal to BA, and in EF make EK equal to BC, join HK.
Then, because AB: BC: DE: EF,
therefore HE: EK:: DE: EF,
and the angle DEF is common to the triangles HEK, DEF,
But the triangle HEK has been proved similar to the triangle DEF,
therefore the triangle ABC is similar to the triangle DEF, having the angles at A, B, C respectively equal to the angles at D, E, F.
THEOR. 5. If two triangles have one angle of the one equal to one angle of the other, and the sides about one other angle in each proportional, so that the sides opposite the equal angles are homologous, the triangles have their third angles equal or supplementary, and in the former case the triangles are similar.
Let ABC, DEF be two triangles having the angle ABC equal to the angle DEF, and the side AB to the side AC as the side DE to the side DF:
then shall the angles ACB, DFE be either equal or supplementary, and, in the former case, the triangles shall be similar.
Apply the triangle ABC to the triangle DEF, so that B may fall on E, and BA along ED, then BC will fall along EF, since the angle ABC is equal to the angle DEF.
Let A and C fall at A' and C' in ED and EF, or in these sides produced.
Then, if A'C' is parallel to DF,
the angle A'C'E is equal to the angle DFE,
that is, the angle ACB is equal to the angle DFE,
But if A'C' is not parallel to DF,
from A' draw A'H parallel to DF, and meeting EF in H.
but EA': A'C':: ED: DF,
therefore EA': A'H:: EA' : A'C',
and therefore A'H is equal to A'C',
therefore the angle A'C'H is equal to the angle A'HC', therefore the angles A'C'E, A'HC' are supplementary, and therefore the angles ACB, DFE are supplementary.
COR. Two such triangles are similar
1. If the two angles given equal are right angles or obtuse angles.
2. If the angles opposite to the other two homologous sides are both acute or both obtuse, or if one of them is a right angle.
3. If the side opposite the given angle in each triangle is not less than the other given side.
Ex. 9. If OMN, OPQ are two straight lines, and MP, NQ meet in R, then if OM MP :: ON: NQ, the triangle PQR is isosceles.
THEOR. 6. If two similar rectilineal figures are placed so as to have their homologous sides parallel, all the straight lines joining the angular points of the one to the corresponding angular points of the other are parallel or meet in a point; and the distances from that point along any straight line to the points where it meets homologous sides of the figures are in the ratio of the homologous sides of the figures.
Let ABCD, HKLM be two similar rectilineal figures having the sides AB, BC, CD, DA respectively parallel to the homologous sides HK, KL, LM, MH:
then shall the straight lines AH, BK, CL, DM be all parallel or meet in a point.
First, let the lines AB, HK be drawn in the same sense (ie., in the same direction, or towards the same parts) from A and H.
Then, if the figures are equal as well as similar, ABKH is a quadrilateral having its opposite sides AB, HK equal and parallel, therefore AH is parallel to BK;
and in like manner BK is parallel to CL,
and CL to DM.
But if the figures are not equal,
let AH be produced to cut BK externally in O.
Then because AB and HK are parallel,
therefore the triangles ABO, HKO are equiangular,
and therefore BO: KO:: AB: HK,
that is, AH cuts BK externally in the ratio AB: HK.
In like manner CL cuts BK externally in the ratio BC: KL, but AB HK:: BC: KL, since the figures are similar, therefore AH and CL cut BK externally in the same ratio, and therefore in the same point O.
By like reasoning, DM passes through the point O in which BK cuts CL.
Hence AH, BK, CL, DM meet in the same point O.
Secondly, let the lines AB, HK be drawn in contrary senses from A and H.