gle ABC which the opposite seg ments BA and BC make with each other being equal to two right angles, the straight line that bisects it must be the perpendicular required. Taking BD, therefore, equal to BE, and construct

F

A D

B

E

ing the isosceles triangle DFE; the straight line BF which joins the vertex of the triangle is perpendicular to AC.

PROP. VI. PROB.

To let fall a perpendicular upon a straight line, from a given point without it.

From the point C to let fall a perpendicular upon a given straight line AB.

In AB take the point D, and with the distance DC describe a circle; and in the same line take another point E, and with distance EC describe a second circle intersecting the former in F; join CF, cross

ing the given line in G: CG is perpen

dicular to AB.

For the triangles DCE and DFE have the side DC equal to DE, CE to FE, and DE common to them both; whence (I. 2.) the angle CDE or CDG

-B

is equal to FDE or FDG. And because in the triangles DCG and DFG, the side DC is equal to DF, DG is common, and the contained angles CDG and FDG are proved to be equal; these triangles are (I. 3.) equal, and consequently the angle DGC is equal to DGF, and each of them a right angle, or CG is perpendicular to AB.

PROP. VII. PROB.

To bisect a given finite straight line.

On the given straight line AB construct two isosceles triangles (I. 1.) ACB and ADB, and join their vertices C and D by a straight line cutting AB in the point E: AB is bisected in E.

E

B

For the sides AC and AD of the triangle CAD being respectively equal to CB and BD of the triangle CBD, and the side CD common to them both; these triangles (I. 2.) are equal, and the angle ACD or ACE is equal to BCD or BCE. Again, the triangles ACE and BCE, having the side AC equal to BC, CE common, and the contained angle ACE equal to BCE, are (I. 3.) equal, and consequently the base AE is equal to BE.

PROP VIII. THEOR.

The angles at the base of an isosceles triangle are equal.

The angles BAC and BCA at the base of the isosceles triangle ABC are equal.

For draw (I. 5.) BD bisecting the vertical angle ABC.

Because AB is equal to BC, the side BD common to the two triangles BDA and BDC, and the angles ABD and CBD contained by them are equal; these triangles are equal (I.

B

3.) and consequently the angle BAD is equal to BCD. Cor. Every equilateral triangle is also equiangular.

PROP. IX. THEOR.

If two angles of a triangle be equal, the sides opposite to them are likewise equal.

Let the triangle ABC have two equal angles BCA and BAC; the opposite sides AB and BC are also equal.

For if AB be not equal to CB, let it be equal to the part CD, and join AD.

Comparing now the triangles BAC and DCA, the side AB is by supposition equal to CD, AC is common to both, and the contained angle BAC is equal to DCA; the two triangles (I. 3.) are, therefore, equal. But this A

conclusion is manifestly absurd. To suppose then the inequality of AB and BC, involves a contradiction; and consequently those sides must be equal.

Cor. Every equiangular triangle is also equilateral.

PROP. X. THEOR.

The exterior angle of a triangle is greater than either of the interior opposite angles.

The exterior angle BCF, formed by producing a side AC of the triangle ABC, is greater than either of the opposite and interior angles CAB and CBA.

For bisect the side BC in D (I. 7.), draw AD, and produce it until DE be equal to AD, and join EC.

The triangles ADB and CDE have by construction the side DA equal to DE, the side DB to

DC, and the vertical angle BDA

B

D

G

is equal to CDE (Def. 10.); these triangles are, therefore,

B

equal (I. 3.), and the angle DCE is equal to DBA. But the angle BCF is evidently greater than DCE, it is consequently greater than DBA or ABC.

In like manner, it may be shown, that if BC be produced, the exterior angle ACG is greater than CAB. But ACG is equal to its vertical angle BCF (Def. 10.), and hence BCF must be greater than either the angle CBA or CAB,

PROP. XI. THEOR.

Any two angles of a triangle are together less than two right angles.

The two angles BAC and BCA of the triangle ABC are together less than two right angles.

For produce the common side AC. And by the last proposition the exterior angle BCD is greater than CAB; add BCA to each, and the two angles BCD and BCA are greater than

CAB and BCA, or CAB and BCA are together less than BCD and BCA, that is, less than two right angles (Def. 4).

PROP. XII. THEOR.

Every triangle has two acute angles.

Let the triangle ABC have first a right angle at C.

Then, by the last proposition, the angles

B

ACB and CAB are less than two right an

gles, and so are the angles ACB and ABC.

Consequently the angles CAB and CBA A

C

are each of them less than one right angle, or they are

both acute.

Next let the triangle have an obtuse angle ACB. The angles ACB and CAB, being less than two right angles, and ACB being greater than one

right angle, CAB must be much less

than a right angle. And the angles

ACB and ABC being also less than two right angles, ABC must be much less A

B

than one right angle. Consequently the angles CAB and CBA are both of them acute.

Lastly, let the triangle have the angle at C acute. If one of the remaining angles, such as BAC,

be likewise acute, the two angles ACB and BAC are both of them acute. But if the angle BAC be either obtuse or a right angle, it comes under the two former cases,

and the other angles ABC and ACB are, therefore, acute,

PROP. XIII. THEOR.

If from a point without a straight line, two other straight lines be drawn to meet it; the nearer one will form on the same side a greater angle than that which is more remote.

If straight lines CD, CE be drawn from the point C to the straight line AB; the angle ADC is greater than AEC.

For ADC is the exterior angle of the triangle DCE, and is consequently (I. 10.) greater than the opposite interior angle CED.

A DE

C

B

If the line CD be, therefore, supposed to turn about the point C in the direction of AB, the angle which it makes with the intercepted part of the line from A will continually diminish.