EU. I. 7. THEOR.

On the same base, and on the same side of it, there cannot be two triangles having 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.

If possible let triangles ACB, ADB on the same side of the common base AB have AC AD and BC = BD.

=

Join CD, and, in the case where the vertex

of the one triangle falls within the other, A
produce AC and AD.

D

1st Fig.. AD = AC (hyp.), .. 4 ADC = 4 ACD (I. 5),

But BD =

B

BC (hyp.), .. ▲ BDC= 4 BCD (I. 5), which is impossible.

.. AC AD and BC = BD cannot both be true.

Note 1.-Proposition 7 is only required by Euclid in the demonstration of proposition 8, and is not used again. The two are superseded in the text by another demonstration of proposition 8, not dependent on proposition 7.

Note 2. The following is a simple way of proving Prop. 7 with the help of a deduction from Prop. 5. It is easily proved (Ex. 8) that if two isosceles triangles be on the same base, the straight line joining their vertices shall bisect the base. Now, in the proposition above, CAD and CBD are supposed isosceles triangles on base CD, therefore CD is bisected by AB, which it is not as C and D are on the same side of AB, and therefore ČAD and CBD are not two isosceles triangles.

EU. I. 8. THEOR.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also 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.

And if ABAC do not coincide with AEDF, let it fall as

AEGF.

Then.. EG = BA, and BA = ED, ... ED = EG;

and in like manner FG=FD; which is impossible (I. 7). .. ABAC must coincide with ▲ EDF,

.. LA = LD. Q.E.D.

COR.-AABC coincides with ÄDEF, and .. ΔΑΒΕ ADEF in every respect.

=

EU. I. 24. THEOR.

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 the one greater than the angle contained by the two sides, equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.

A.

Let triangles ABC and DEF have AB = DE, AC = DF, and A> EDF.

To prove BC > EF.

"Let DE be the side which

is not greater than the B
other."

At D in ED make EDG = ▲ BAC (I. 23),
and make DG =

(cons.), ... AEDG

Again,

and.

DF = AC (I. 3). Join EG and GF.
AB (hyp.), DG = AC (cons.), and ▲ EDG = ▲ A

=

DG DF (cons.), .. ▲ DGF = ≤ DFG (I. 5),
DGF is > EGF, ... DFG is > EGF,
EFG > < EGF,

much more then is

and.. EG is >EF (I. 19). But EG

=

BC,..
.. BC >EF. Q.E.D.

Note. Without the proviso marked with inverted commas, which was added by Dr Simson, the point F could occupy three positions, viz., on, above, or below EG, making three cases of the proposition. With the help of Simson's condition it can be proved that F falls below EG.

Let DF and EG meet at H.

... ▲ DHG >▲ DEG (I. 16), and DEG is not less than ▲ DGE (I. 18),

.. ▲ DHG>▲ DGH, ... DG is>DH (I. 19), .'. DF>DH. .. F falls below EG.

Assuredly the proposition is not complete without a proof of this fact.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also 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.

And if ABAC do not coincide with AEDF, let it fall as

AEGF.

Then. EG BA, and BA

=

=

ED,... ED = EG;

and in like manner FG=FD; which is impossible (I. 7). .. ABAC must coincide with AEDF,

.. LA = 2 D. Q.E.D.

COR.-AABC coincides with ADEF, and ... AABC = ADEF in every respect.

=

EU. I. 24. THEOR.

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 the one greater than the angle contained by the two sides, equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.

At D in ED make EDG = BAC (I. 23),

and make DG == DF =

AC (I. 3). Join EG and GF.
AC (cons.), and ▲ EDG

LA

DFG (I. 5),

ABAC (I. 4), and ... EG = BC.

=

DF (cons.), ... 4 DGF

=

and. DGF is > ▲ EGF, ... ≤ DFG is > < EGF, much more then is EFG > < EGF,

and.. EG is >EF (I. 19).

But EG =

BC,.. BC >EF. Q.E.D.

Note.-Without the proviso marked with inverted commas, which was added by Dr Simson, the point F could occupy three positions, viz., on, above, or below EG, making three cases of the proposition. With the help of Simson's condition it can be proved that F falls below EG.

Let DF and EG meet at H.

·.· ▲ DHG >▲ DEG (I. 16), and ▲ DEG is not less than ▲ DGE (I. 18),

.. 4 DHG>4 DGH,... DG is >DH (I. 19), .'. DF>DH. .. F falls below EG.

Assuredly the proposition is not complete without a proof of this fact.