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NOTE. It is shown in Optics that rays of light, incident from B on a plane mirror through CD at right angles to the plane of the paper, proceed after reflection as if they came from F. On this account F is sometimes called the Image of B in CD.

2. If, from two fixed points on opposite sides of a given straight line straight lines be drawn to a point in the line, the difference of the lengths of these lines shall be greatest when they make equal angles with the given line.

[blocks in formation]

Draw BECD, and produce it to meet AP in F.
Show as in § 6, Ex. 7 that AQ-QF<AF,

or, AQ-QB<AP-PB.

3. Of all triangles on the same base and having the same area, the isosceles has the least perimeter.

A

B

Let AB AC and AD || BC.
Then ADBC=▲ ABC.

[Euc. I. 37.

Show that AB, AC make equal angles with AD,

and apply Ex. 1.

4. Of all triangles which have the same vertical angle, and whose bases pass through a fixed point between the arms of the angle, the least is that whose base is bisected at the fixed point. A

B

Let P be the fixed point, and let BC be bisected at P, and

DPE be any other base.

Draw CFDB,

and use Euc. I. 26, etc.

5. From a point in the base of a triangle lines are drawn parallel to the sides. Show that the parallelogram which is thus formed is greatest when the point bisects the base.

R

Let BC be bisected in D.
Use § 16, Ex. 2 to show
AD=ABC.

If QPR be drawn through P,
so that QPPR,

show APA AQR;
and show as in Ex. 4 that

▲ AQRA ABC,

and..

AP<OAD.

6. Of all triangles having the vertical angle and the sum of the sides the same, the isosceles triangle has the smallest base.

Let ABC be an isosceles ▲, having

AB=AC.

Let AP+AQ=AB+AC.

Draw QR and || PB.

Show that CQR is isosceles and CRL BC, and.. BR (or PQ) > BC.

7. Of all triangles having two given sides, the greatest is that

in which the contained angle is a right angle.

Let AB, BC be the given sides, and

let ABC be a right 4.

[blocks in formation]

8. Of all triangles having the same base and the same perimeter, the isosceles has the greatest area.

Let AB=BC,

and AB+BC=AD+DC.

E

[Ex. 1.

Let BE be drawn || AC meeting AD in E.
Then AE+EC>AB+BC

>AD+DC;

D is between A and E.

Hence show A ABC>^ ADC.

9. If the diagonals of a quadrilateral be given in magnitude, its area shall be a maximum when they are at right angles. Use the methods of Ex. 7 and § 16, Ex. 17.

10. If the diagonals of a parallelogram be given, its area shall be a maximum when it is a rhombus.

11. If two sides of a triangle be unequal, the altitude drawn to the shorter side shall be greater than that drawn to the longer.

Use § 18, Ex. 5.

12. XOY is an acute angle, and A, B are points within it. If P, Q be points on OX, OY, show that AP+PQ+QB is a minimum when APX= < OPQ and ▲ PQO= < BQY.

Use Ex. 1.

13. If, in the figures of Exx. 1 and 2, AB meet CD in R, show that AR-BR and AR+BR are respectively a maximum and a minimum.

CHAPTER II.

PROBLEMS.

THE solution of problems often presents more difficulty than the proof of theorems, as the enunciation may give little or no clue to the construction required. It will, however, be found that nearly all problems may be solved by one or other of the methods which are given in the following sections, and that many problems may be solved by more than one of these methods. The special attention of the student should be given to the method of Analysis and Synthesis described in § 27.

The remarks in Italics, which in the illustrative examples follow the general enunciations, are intended to explain to the student the nature of the method employed in the section, and do not require to be written out as part of the solution.

$22. Problems which follow directly from known Theorems. 1. Trisect a right angle.

We know by Euc. I. 32 that each angle of an equilateral triangle is two-thirds of a right angle. This suggests the construction which follows. (A Standard Construction.)

E

B

Let ABC be a right angle; it is required to trisect it.

86

Construction—

On BC construct an equilateral ▲ DBC.
Through B draw BE bisecting the DBC.
LABC shall be trisected by BD and BE.

Proof

[Euc. I. 1.

[Euc. I. 9.

The sum of the three angles of ADBC is 2 rt. 4 s. [Euc. I. 32. .. 4 DBC is two-thirds of a rt. ;

.. each of the s ABD, DBE, EBC is one-third of a rt. 4.

2. Bisect a parallelogram by a straight line drawn through any given point.

[blocks in formation]

3. Divide an equilateral triangle into three congruent triangles.

Use § 9, Ex. 6, and Euc. I. 26.

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