« ΠροηγούμενηΣυνέχεια »
PROPOSITION B. THEOREM.
If two triangles have two angles of the one equal to two angles of the other, each to each, and the sides adjacent to the equal angles in each also equal; then must the triangles be equal in all respects.
let ABC= DEF, and ACB= 4 DFE, and BC=EF. Then must AB=DE, and AC=DF, and ▲ BAC=1 EDF. For if A DEF be applied to AABC, so that E coincides with B, and EF falls on BC;
then. EF=BC, .. F will coincide with C;
and ▲ DEF= 2 ABC, .. ED will fall on BA ;
Again, . ▲ DFE= ▲ ACB, .. FD will fall on CA ;
.. D must coincide with A, the only pt. common to BA and CA.
.. DE will coincide with and .. is equal to AB,
COR. Hence, by a process like that in Prop. A, we can prove the following theorem:
If two angles of a triangle be equal, the sides which subtend them are also equal. (Eucl. I. 6.)
PROPOSITION C. THEOREM.
If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles must be equas in all respects.
Let the three sides of the As ABC, DEF be equal, each to each, that is, AB=DE, AC=DF, and BC=EF.
Then must the triangles be equal in all respects.
Imagine the ▲ DEF to be turned over and applied to the ▲ ABC, in such a way that EF coincides with BC, and the vertex D falls on the side of BC opposite to the side on which A falls; and join AD.
CASE I. When AD passes through BC.
Then in ▲ ABD, ·· BD=BA, .. 2 BAD= 2 BDA, I. A. And in ▲ ACD, ::: CD=CA,.. 4 CAD= 2 CDA, I. A. ... sum of 48 BAD, CAD=sum of 1 s BDA, CDA, Ax. 2. that is, L BACL BDC.
Hence we see, referring to the original triangles, that
.., by Prop. 4, the triangles are equal in all respects.
CASE II. When the line joining the vertices does not pass through BC.
Then in ▲ ABD, ·· BD=BA, :. ▲ BAD= L BDA, I. A.
and parts of these CAD, CDA are equal.
Then, as in Case I., the equality of the original triangles may be proved.
CASE III. When AC and CD are in the same straight line.
Then in ▲ ABD, ·· BD=BA, .. ▲ BAD≈ ▲ BDA, I. A.
Then, as in Case I., the equality of the original triangles
may be proved.
Q. E. D.
Let the three sides to each, that is, AB
Then must the t
A falls; a
e the given angle.
required to bisect & BAC.
any pt. D.
In AC make AE=AD, and join DE.
On DE, quilat. ADFE
Join AF. Then AF will bisect ▲ BAC.
For in As AFD, AFE,
::AD=AE, and AF is common, and FD=FE,
.. ▲ DAF= LEAF,
that is, BAC is bisected by AF.
Q. E. F.
Ex. 1. Shew that we can prove this Proposition by means of Prop. IV. and PROP. A., without applying Prop. C.
Ex. 2. If the equilateral triangle, employed in the construction, be described with its vertex towards the given angle; shew that there is one case in which the construction will fail, and two in which it will hold good.
NOTE.-The line dividing an angle into two equal parts is led the BISECTOR of the angle.
Bisect ▲ ACB by the st. line CD meeting AB in D ̧
then AB shall be bisected in D.
For in as ACD, BCD,
:: AC=BC, and CD is common, and ▲ ACD= ▲ BCD,
.. AD=BD ;
.. AB is bisected in D.
Q. E. F.
Ex. 1. The straight line, drawn to bisect the vertical angle of an isosceles triangle, also bisects the base.
Ex. 2. The straight line, drawn from the vertex of an isosceles triangle to bisect the base, also bisects the vertical angle.
Ex. 3. Produce a given finite straight line to a point, such that the part produced may be one-third of the line, which is made up of the whole and the part produced.