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PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of every triangle are together equal to two right angles.

[blocks in formation]

II.

LS ABC, BAC, ACB together=two rt. 4 s.

From C draw CE || to AB.

I. 31.

Then I... BD meets the Is EC, AB,
.. extr. 4 ECD=intr. ▲ ABC.

I. 29.

And AC meets the Is EC, AB,

.. ACE alternate BAC.

I. 29.

.. 48 ECD, ACE togethers ABC, BAC together; .. ▲ ACD= Ls ABC, BAC together.

And II. ... LS ABC, BAC together

ACD,

to each of these equals add 4 ACB;

thens ABC, BAC, ACB together= 2 s ACD, ACB together,

.. Ls ABC, BAC, ACB together=two rt. 4 s.

I. 13.

Q. E. D.

Ex. 1. In an acute-angled triangle, any two angles are greater than the third.

Ex. 2. The straight line, which bisects the external vertical angle of an isosceles triangle is parallel to the base.

Ex. 3. If the side BC of the triangle ABC be produced to D, and AE be drawn bisecting the angle BAC and meeting BC in E; shew that the angles ABD, ACD are together . double of the angle AED.

Ex. 4. If the straight lines bisecting the angles at the base of an isosceles triangle be produced to meet; shew that they will contain an angle equal to an exterior angle at the base of the triangle.

Ex. 5. If the straight line bisecting the external angle of a triangle be parallel to the base; prove that the triangle is isosceles.

The following Corollaries to Prop. 32 were first given in Simson's Edition of Euclid.

COR. 1. The sum of the interior angles of any rectilinear figure together with four right angles is equal to twice as many right angles as the figure has sides.

B

A

E

Let ABCDE be any rectilinear figure.

Take any pt. F within the figure, and from F draw the st. lines FA, FB, FC, FD, FE to the angular pts. of the figure Then there are formed as many s as the figure has sides.

The threes in each of these As together two rt. 4 s.

..all the 4s in these As together twice as many right ▲s as there are ▲s, that is, twice as many rights as the figure has sides.

Now angles of all the Ass at A, B, C, D, E and ▲ s

at F,

that is,

and..

= 48 of the figure and ▲ s at F,

= 48 of the figure and four rt. 4 s. I. 15. Cor. 2. ..s of the figure and four rt. 4 s=twice as many rt. 48 as the figure has sides.

COR, 2. The exterior angles of any convex rectilinear figure, made by producing each of its sides in succession, are together equal to four right angles.

Every interior angle, as ABC, and its adjacent exterior angle, as ABD, together are two rt. 4 s.

D B

.. all the intr. 4s together with all the extr. 4 s
=twice as many rt. 4s as the figure has sides.
But all the intr. 4s together with four rt. s
twice as many rt. 4 s as the figure has sides.

..all the intr. 4s together with all the extr. 4s =all the intr. 4s together with four rt. 4 s.

.. all the extr. 4 s=four rt. 4 s.

NOTE. The latter of these corollaries refers only to convex figures, that is, figures in which every interior angle is less than two right angles. When a figure contains an angle greater

than two right angles, as the angle marked by the dotted line in the diagram, this is called a reflex angle. See p. 149.

Ex. 1. The exterior angles of a quadrilateral made by producing the sides successively are together equal to the interior angles.

Ex. 2. Prove that the interior angles of a hexagon are equal to eight right angles.

Ex. 3. Shew that the angle of an equiangular pentagon is of a right angle.

Ex. 4. How many sides has the rectilinear figure, the sum of whose interior angles is double that of its exterior angles? Ex. 5. How many sides has an equiangular polygon, four of whose angles are together equal to seven right angles?

PROPOSITION XXXIII. THEOREM.

The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.

Let the equal and || st. lines AB, CD be joined towards the same parts by the st. lines AC, BD.

Then must AC and BD be equal and \\.

[blocks in formation]

::* AB=CD, and BC is common, and ▲ ABC= ▲ DCB,

L

.. AC-BD, and ▲ ACB= DBC.

Then BC, meeting AC and BD,

s ACB, DBC equal,

makes the alternate

.. AC is to BD.

I. 29.

I. 4.

Q. E. D.

Miscellaneous Exercises on Sections I. and II.

1. If two exterior angles of a triangle be bisected by straight lines which meet in O; prove that the perpendiculars from O on the sides, or the sides produced, of the triangle are equal.

2. Trisect a right angle.

3. The bisectors of the three angles of a triangle meet in one point.

4. The perpendiculars to the three sides of a triangle drawn from the middle points of the sides meet in one point.

5. The angle between the bisector of the angle BAC of the triangle ABC and the perpendicular from A on BC, is equal to half the difference between the angles at B and C.

6. If the straight line AD bisect the angle at A of the triangle ABC, and BDE be drawn perpendicular to AD, and meeting AC, or AC produced, in E; shew that BD is equal to DE.

7. Divide a right-angled triangle into two isosceles triangles.

8. AB, CD are two given straight lines. Through a point E between them draw a straight line GEH, such that the intercepted portion GH shall be bisected in E.

9. The vertical angle O of a triangle OPQ is a right, acute, or obtuse angle, according as OR, the line bisecting PQ, is equal to, greater or less than the half of PQ.

10. Shew by means of Ex. 9 how to draw a perpendicular to a given straight line from its extremity without producing it.

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