PROBLEM N.

Describe a triangle equal to a given rectilineal figure.

Let PQRSTV be the given rectilineal figure.
It is required to describe a ▲ equal to it.

Join PT: and draw VH || to PT, meeting ST produced

in H.

Join PH.

Then ▲ PTH is =▲ PVT;

add to each the figure PQRST;

.. figure PQRSH is

=

the figure PQRSTV.

(1. 32)

Similarly we may obtain a figure = RH having one fewer sides, and so on; thus we may obtain a ▲ equal to the given figure.

COR. Hence a parallelogram may be drawn = a given rectilineal figure and having an angle = a given angle.

SUPPLEMENT TO BOOK I.

THEOREM (a).

Of all straight lines which can be drawn to a given straight line from a given point without it, that which is nearer to the perpendicular is less than the more remote; also to every straight line drawn on one side of the perpendicular there can be drawn one and only one straight line equal to it on the other side.

Let AP be the perpendicular from A on BC; AQ, AR other straight lines on the same side of AP.

Also if the figure APQRB revolve about AP,

then PB will fall on PC and Q, R on points Q', R' in the line PC,

and .. AQ, AR on AQ, AR';

so that to every straight line on one side of AP an equal straight line can be drawn on the other side ;

and no two straight lines on the same side of the perpendicular can be equal;

.. there cannot be drawn from it more than two straight lines equal to one another.

THEOREM (b).

If in any two sides and an opposite to one of them are respectively equal to the corresponding sides and LS in another A, then shall these as be equal in certain cases.

Let ABC, DEF be two as having

the sides BA, AC and ABC respectively

then shall ▲ ABC be = A DEF in certain cases.

Let the ▲ ABC be applied to the ▲ DEF so that AB may be on DE and BC along EF.

Then if a DFE is a right no other straight line can be drawn from D=DF;

.. AC must fall on DF

Hence ▲ ABC is ▲ DEF in all respects.

=

(I. 12)

But if DFE is not a right, then one other straight line DF', and only one, can be drawn = DF;

.. AC must fall on DF or DF'.

Hence, if the s opposite the sides AB, DE are both acute or both obtuse,

then will the as ABC, DEF be equal in all respects.

DEFINITION.

The locus of a point satisfying a certain geometrical condition is the line or lines composed of all possible positions of that point.

PROBLEM (c).

Find the locus of a point equidistant from two given points.

.. Q lies on the line bisecting AB at right ≤ s.

Similarly every other point equidistant from A and B may be shewn to lie on the line bisecting AB at rights; i.e. the locus consists of points which lie on this line.

Moreover, every point in the line bisecting AB at right angles is a point of the locus.

For if any point R be taken in this line, then AP, PR, and the ▲ APR are respectively = BP, PR, and the ▲ BPR ; .. R is equidistant from A and B and is,. a point of the locus.

Hence the locus of the point equidistant from A and B. is the straight line bisecting AB at rights.

DEDUCTION (d).

The lines bisecting the sides of a ▲ at rights meet in a

point.

H

F

B

Let ABC be a ▲, bisect AC, BC in H, K,

and draw HF 1 AC and KF 1 BC.

Let these perpendiculars meet in F.

Then HF is the locus of points equidistant from A and C; .. Fis equidistant from A and C.

Similarly Fis equidistant from B and C ;

.. Fis equidistant from A and B;

.. Flies on the line bisecting AB at rights;

.. the line bisecting AB at right 4 s passes through F;

Δ

.. the lines bisecting at right 4 s the sides of the a ABC meet in the point F

C. G.

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