PROPOSITION XXIV. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other; the base of that which has the greater angle must be greater than the base of the other.

In the As ABC, DEF,

let AB=DE and AC=DF,

and let ▲ BAC be greater than ▲ EDF.

Then must BC be greater than EF.

Of the two sides DE, DF let DE be not greater than DF.* At pt. D in st. line ED make ▲ EDG= 2 BAC,

and make DG=AC or DF, and join EG, GF.

L

I. 23.

Then·.· AB=DE, and AC=DG, and ▲ BAC= ▲ EDG,

much more then ▲ EFG is greater than EGF; .. EG is greater than EF.

But EG=BC;

.. BC is greater than EF.

I. 19.

Q. E. D.

*This line was added by Simson to obviate a defect in Euclid's Without this condition, three distinct cases must be discussed. With the condition, we can prove that F must lie below EG.

proof.

For since DF is not less than DE, and DG is drawn equal to DF, DG is not less than DE.

Hence by Prop. D, any line drawn from D to meet EG is less than DG, and therefore DF, being equal to DG, must extend beyond EG.

For another method of proving the Proposition, see p. 113.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle also, contained by the sides of that which has the greater base, must be greater than the angle contained by the sides equal to them of the other.

E

In the As ABC, DEF,

let AB DE and AC=DF,

and let BC be greater than EF.

Then must BAC be greater than

EDF.

For BAC is greater than, equal to, or less than EDF. BAC cannot ▲ EDF,

Now

for then, by 1. 4, BC would=EF; which is not the case. And 4 BAC cannot be less than ▲ EDF,

for then, by 1. 24, BC would be less than EF; which is not the case;

.. 4 BAC must be greater than EDF.

Q. E. D.

NOTE.-In Prop. xxvI. Euclid includes two cases, in which two triangles are equal in all respects; viz., when the following parts are equal in the two triangles :

1. Two angles and the side between them.

2. Two angles and the side opposite one of them.

Of these we have already proved the first case, in Prop. B, so that we have only the second case left, to form the subject of Prop. XXVI., which we shall prove by the method of

superposition.

For Euclid's proof of Prop. XXVI., see rn. 114-115.

PROPOSITION XXVI. THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, those sides being opposite to equal angles in each; then must the triangles be equal in all respects.

let ABC= DEF, and ▲ ACB= ▲ DFE, and AB=DE. Then must BCEF, and AC=DF, and ▲ BAC= ▲ EDF. Suppose ▲ DEF to be applied to ▲ ABC,

so that D coincides with A, and DE falls on AB.

Then.DE=AB,... E will coincide with B ;

and. ▲ DEF= 2 ABC,.. EF will fall on BC. Then must F coincide with C: for, if not,

let F fall between B and C, at the pt. H. Join AH.

Then

:: 2 AHB= 2 DFE,

.. 2 AHB= 2 ACB,

I. 4.

the extr. the intr. and opposite ▲, which is impossible. .. F does not fall between B and C.

Similarly, it may be shewn that F does not fall on BC produced.

.. F coincides with C, and .. BC=EF ;
.. AC=DF, and ▲ BAC=▲ EDF,
and.. the triangles are equal in all respects.

I. 4.

Q. E. D.

Miscellaneous Exercises on Props. I. to XXVI.

1. M is the middle point of the base BC of an isosceles triangle ABC, and N is a point in AC. Shew that the difference between MB and MN is less than that between AB and AN.

2. ABC is a triangle, and the angle at A is bisected by a straight line which meets BC at D; shew that BA is greater than BD, and CA greater than CD.

3. AB, AC are straight lines meeting in A, and D is a given point. Draw through D a straight line cutting off equal parts from AB, AC.

4. Draw a straight line through a given point, to make equal angles with two given straight lines which meet.

5. A given angle BAC is bisected; if CA be produced to G and the angle BAG bisected, the two bisecting lines are at right angles.

6. Two straight lines are drawn to the base of a triangle from the vertex, one bisecting the vertical angle, and the other bisecting the base. Prove that the latter is the greater of the two lines.

7. Shew that Prop. XVII. may be proved without producing a side of the triangle.

8. Shew that Prop. XVIII. may be proved by means of the following construction: cut off AD=AB, draw AE, bisecting L BAC and meeting BC in E, and join DE.

9. Shew that Prop. xx. can be proved, without producing one of the sides of the triangle, by bisecting one of the angles.

10. Given two angles of a triangle and the side adjacent to them, construct the triangle.

11. Shew that the perpendiculars, let fall on two sides of a triangle from any point in the straight line bisecting the angle contained by the two sides, are equal.

We conclude Section I. with the proof (omitted by Euclid) of another case in which two triangles are equal in all respects.

PROPOSITION E. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle in each equal: then, if the third angles in each be both acute, both obtuse, or if one of them be a right angle, the triangles are equal in all respects.

AA

In the As ABC, DEF, let ▲ BAC= ▲ EDF, AB=DE, BC=EF, and let s ACB, DFE be both acute, both obtuse, or let one of them be a right angle.

Then must as ABC, DEF be equal in all respects.

For if AC be not =DF, make AG=DF; and join BG.
Then in AS BAG, EDF,

·.· BA=ED, and AG=DF, and ▲ BAG= ▲ EDF,
.. BG EF and ▲ AGB= ▲ DFE.

But BC=EF, and .. BG=BC;

.. 4 BCG= L BGC.

First, let

ACB and ▲ DFE be both acute,

I. 4.

I. A.

I. 13.

then AGB is acute, and .. 4 BGC is obtuse;

.. BCG is obtuse, which is contrary to the hypothesis. Next, let

ACB and DFE be both obtuse,

then AGB is obtuse, and .. 4 BGC is acute; .. BCG is acute, which is contrary to the hypothesis.

I. 13.