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

=

In As ABC, DEF,

DEF, and 4 ACB= ▲ DFE, and AB=DE.

Then must BC=EF, and AC=DF, and ▲ BAC= ▲ EDF. Suppose DEF to be applied to ▲ ABC,

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

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

and

DEF = L 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

:: LAHB= ▲ DFE,

.. LAHB= L ACB,

I. 4.

the extr. 4 =

the intr. and opposite 4, 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 ;

EDF,

.. AC=DF, and ▲ BAC
and... the triangles are equal in all respects.

I. 4.

Q. E. D.

Book I.]

MISCELLANEOUS EXERCISES.

41

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 ▲ 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.

E

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= L DFE.

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

.. 4 BCGL BGC.

First, let

ACB and ▲ DFE be both acute,

then

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

I. 4.

I. A.

I. 13.

.. 4 BCG is obtuse, which is contrary to the hypothesis.

Next, let

ACB and 4 DFE be both obtuse,

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

I. 13.

Lastly, let one of the third angles ACB, DFE be a right

angle.

If ACB be a rt. 4,

then 4 BGC is also a rt. 4;

I. A.

.. 48 BCG, BGC together two rt. 4s, which is impossible. Again, if DFE be a rt. 4,

I. 17.

then

Hence

AGB is a rt. 4, and .. ▲ BGC is a rt. 4.
BCG is also a rt. 4.

I. 13.

.. 48 BCG, BGC together=two rt. ▲ s, which is impossible.

Hence AC is equal to DF,

and the AS ABC, DEF are equal in all respects.

I. 17.

Q. E. D.

COR. From the first case of this proposition we deduce the following important theorem :

If two right-angled triangles have the hypotenuse and one side of the one equal respectively to the hypotenuse and one side of the other, the triangles are equal in all respects.

NOTE. In the enunciation of Prop. E, if, instead of the words if one of them be a right angle, we put the words both right angles, this case of the proposition would be identical with I. 26.