PROBLEM 4.

To draw a straight line perpendicular to a given straight line from a given external point.

Let X be the given external point from which a perpendicular is to be drawn to AB.

Construction. Take any point C on the side of AB remote from X.

With centre X, and radius XC, draw an arc to cut AB at P and Q.

With centres P and Q, and radius PX, draw arcs cutting at Y, on the side of AB opposite to X.

Proof.

Join XY cutting AB at O.
Then XO is perp. to AB.

Join PX, QX, PY, QY,

In the ▲ PXY, QXY,

PX = QX, being radii of a circle,

because PYQY, for the same reason,

and XY is common;

And these being adjacent angles, each is a right angle,

that is, XO is perp. to AB.

Obs. When the point X is nearly opposite one end of AB, one or other of the alternative constructions given below should be used.

PROBLEM 4. SECOND METHOD.

Construction. Take any point D in AB. Join DX, and bisect it at C.

With centre C, and radius CX, draw a circle cutting AB at D and O.

Join XO.

Then XO is perp. to AB.

For, as in Problem 3, Second Method, the XOD is a right

angle.

by Theorem 4, so that the adjacent ▲ DOX, DOY are equal.

That is, XO is perp. to AB.

PROBLEM 5.

At a given point in a given straight line to make an angle equal to a given angle.

ДД

A

D B

FO

Q G

Let BAC be the given angle, and FG the given straight line; and let O be the point at which an angle is to be made equal

to the BAC.

Construction.

With centre A, and with any radius, draw

an arc cutting AB and AC at D and E.

With centre O, and with the same radius, draw an arc cutting FG at Q.

With centre Q, and with radius DE, draw an arc cutting the former arc at P.

Proof.

Join OP.

Then POQ is the required angle.

Join ED, PQ.

In the ▲ POQ, EAD,

OP= AE, being radii of equal circles,
because OQ= AD, for the same reason,
PQED, by construction;

.. the triangles are equal in all respects;
so that the POQ= the LEAD.

Theor. 7.

PROBLEM 6.

Through a given point to draw a straight line parallel to a given straight line.

X

Let XY be the given straight line, and O the given point, through which a straight line is to be drawn par to XY.

Construction. In XY take any point A, and join OA. Using the construction of Problem 5, at the point in the line AO make the AOP equal to the OAY and alternate to it.

Then OP is parallel to XY.

Proof. Because AO, meeting the straight lines OP, XY, makes the alternate L POA, OAY equal;

.. OP is par1 to XY.

The constructions of Problems 3, 4, and 6 are not usually followed in practical applications. Parallels and perpendiculars may be more quickly drawn by the aid of set squares. (See LESSONS IN EXPERIMENTAL GEOMETRY, pp. 36, 42.)

PROBLEM 7.

To divide a given straight line into any number of equal parts.

Let AB be the given straight line, and suppose it is required to divide it into five equal parts.

Construction. From A draw AC, a straight line of unlimited length, making any angle with AB.

From AC mark off five equal parts of any length, AP, PQ, QR, RS, ST.

Join TB; and through P, Q, R, S draw par1s to TB, meeting AB in p, q, r, s.

Then since the par's Pp, Qq, Rr, Ss, TB cut off five equal parts from AT, they also cut off five equal parts from AB. (Theorem 22.)

SECOND METHOD.

From A draw AC at any angle with AB, and on it mark off four equal parts AP, PQ, QR, RS, of any length.

From B draw BD par to AC, and on it mark off BS', S'R', R'Q', Q'P', each equal to the parts marked on AC.

Join PP, QQ, RR', SS' meeting AB
in p, q, r, s.
Then AB is divided into

five equal parts at these points.
[Prove by Theorems 20 and 22.]