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the hypothesis of the other. In prop. 5 the hypothesis is the equality of the sides, and the conclusion is the equality of the base angles. In prop. 6 the hypothesis is the equality of the base angles, and the conclusion the equality of the sides. Converse propositions are not universally true : an example of this will be given in the note to prop. 8. This is the first proposition in which we find an indirect (see note before prop. 1) proof, and it may be observed that few converse propositions admit of a direct proof.]
The angles at the base of an isosceles triangle are bisected : show that the bisecting lines form with the base another isosceles triangle.
PROPOSITION VII. THEOREM.
On the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another.
If it be possible, on the same base AB, and on the same side of it, let there be two As ACB, ADB, which have their sides CA, DA, terminated at the extremity A of the base, equal to one another, and likewise their sides CB, DB, that are terminated at B, equal to one another.
First. When the vertex of each of the As is without the other A. Constr.
Join CD. Dem. .: AC = AD .. LACD = LADC ; (i. 5) but _ACD is greater than _ BCD:
(ax. 9) .. also _ ADC is greater than _ BCD; much more then is _BDC greater than _ BCD. Again, :: BC = BD, (hyp.) :: LBDC= _ BCD;
(i. 5) but _ BDC was proved greater than _ BCD, hence LBDC is both equal to, and greater than _ BCD;
which is impossible.
Secondly. Let the vertex D of A ADB fall within ACB.
Constr. Join CD, produce AC to E, and AD to F.
on the other side of the base CD are equal to one another;
(i. 5) but ZECD is greater than _ BCD;
(ax, 9) .. also FDC is greater than _ BCD; much more then is _ BDC greater than _ BCD.
Again, :: BC= BD .. _ BDC= _ BCD; (i. 5) wherefore BDC is both equal to and greater than
BCD; which is impossible.
Thirdly. The case in which the vertex of one triangle is on a side of the other needs no demonstration.
Therefore, on the same base, &c. Q. E. D. [Hypothesis : two triangles on the same base and on the same side of it. Conclusion : they cannot have their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity. This proposition is required directly only in the proof of prop. 8; indirectly however it is often required, inasmuch as prop. 8 (which depends on it) is frequently made use of.]
If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other.
Let ABC, DEF, be two As, having the two sides AB, AC= the two sides DE, DF, each to each, namely
, AB to DE, and AC to DF, and also the base BC= base EF.
Then shall Z BAC = LEDF.
Dem. For if A ABC be applied to A DEF, so that pt. B
be on E, and st. line BC on EF ; then : BC=EF, (hyp.) :: C shall coincide with F; wherefore, BC coinciding with EF, BA and AC shall
coincide with ED, DF; for, if the base BC coincide with the base EF, but the
sides BA, AC do not coincide with the sides ED, DF,
but have a different situation as EG, FG; Then, on the same base and on the same side of it
there can be two As which have their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity ; but this is impossible. (i. 7)
Therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides
ED, DF; wherefore also _BAC coincides with _EDF, and is equal to it. Therefore, if two triangles, &c.
Q. E. D. [Hypothesis, two triangles having two sides of the one equal to two sides of the other, and also their bases equal. Conclusion, the angle contained by the two
of the one shall be equal to the angle contained by the two sides of the other. The As having been proved to coincide are manifestly equal in all respects. Observe however this is not stated in the conclusion, and in subsequent propositions Euclid uses prop. 4 when he requires more than this. The converse proposition would run thus : 'If two As have the 3 Zs in the one equal to the 3 Zs in the other, each to each, the 3 sides shall be equal, each to each. The beginner may at once satisfy himself that this is not true by looking at this figure]:
1. From every point of a given line the lines drawn to each of two given points on opposite sides of the line are equal ; prove that the line joining the given points will be bisected by the given line at right angles.
2. If two circles cut each other, the line joining their points of intersection is bisected at right angles by the line joining their centres.