PROBLEM (e).

Find the locus of a point equidistant from two straight lines which cut one another.

Let AB, AC be the two given straight lines,
Pa point equidistant from them.

From P let fall PH, PK perpendiculars on AB, AC.
Join AP.

HP,PA, and ▲ AHP are respectively=KF, PA, and ▲ AKP, and s HAP, KAP are both acute;

BAC.

... Plies on the line bisecting

Similarly it may be shewn that every other point which is equidistant from AB, AC, lies on the line bisecting ‹ BAC Also every point in the line bisecting the BAC is equidistant from AB and AC.

For if any point R be taken in this line, then letting fall RS, RT on AB, AC, we have

and AR is common to the two a ́s ARS, ART;

.. RS is RT, or R is equidistant from AB and AC; .. the locus of the point equidistant from AB, AC, is

the line bisecting the angle between them.

DEDUCTION (ƒ).

The bisectors of the Ls of a meet in a point.

B

H

Let ABC be a ▲,

and bisect the 4s ABC, ACB by the straight lines BH, CH intersecting in H.

Then BH is the locus of points equidistant from AB, BC; .. His equally distant from AB and BC.

Similarly it is equally distant from BC and CA,
and is .. equally distant from AB and AC,

and.. lies on the line bisecting the BAC;

... the lines bisecting the 4s of the ▲ ABC meet in the point H.

PROBLEM (g).

To describe a triangle having two of its sides equal to two

Let A, B be the given straight lines and C the given angle.

Draw PQR LC,

=

cut off QSA and QT=B; join ST.

Then QST is the triangle required.

PROBLEM ().

To describe a triangle having two of its angles equal to two given angles, and their common arm equal to a given straight line.

Let A, B be the given angles and C the given straight line. Take PQ C, at P and Q make 4 s QPR, PQR = A and B.

Then PQR is the required ▲.

N. B. This problem is impossible if the two given angles are not together less than two rights.

PROBLEM (k).

To describe a triangle having two of its Ls equal to two given angles, and the side opposite to one of them equal to a given straight line.

Let PAQ and B be the two givens, and C the given straight line;

produce PA to R and make 4 RAS = LB.

Hence 4 QAS must be = the remaining of the required ▲ . Take XY C, and make s YXZ, XYZ= 4 s QAP, QAS.

=

Then XYZ is the required a.

N.B. This problem is impossible if the two given angles are not together less than two right angles.