PROBLEM D.

Describe a triangle having its sides equal to three given straight lines, any two of which are together greater than the third.

Let A, B, C be the three given straight lines, any two of them being together greater than the third.

It is required to describe a ▲ having its sides equal to A, B, C respectively.

Take a straight line DK unlimited towards K; cut off DE, EF, FM equal to A, B, C respectively.

With E as a centre and radius ED describe the circle DGH;

and with F as a centre and radius FM describe the circle LHM.

Join EH, FH.

Then EFH is the ▲ required.

For its sides HE, EF, FH are respectively equal to DE, EF, FM;

and therefore to A, B, C

PROBLEM E.

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

It is required at the point D in DE to make an ▲ = L BAC.

With centre A describe a circle cutting AB, AC in F and G; and with centre D describe an equal circle HKM cutting DE in H.

Join FG; with centre H and radius = FG describe a circle cutting the circle HKM in K.

Join DK; then EDK is the required.

Join HK:

Then, ·.· HD, DK, KH are respectively equal to FA, AG, GF;

.. AS HDK, FAG are equal in all respects; (1.5) and ... HDK is = L FAG.

PROBLEM F.

Find the shortest path from a given point to a given straight line.

B

Let A be the given point and BC the given straight line.

From A let fall AQL to BC.

Then shall AQ be less than any other straight line AR drawn from A to BC.

Hence AQ is the shortest path from A to BC.

NOTE. By the distance of a point from a straight line is meant the shortest path from the point to the line.

Hence the distance of a point from a straight line is the perpendicular let fall upon it from the given point.

DEDUCTION G.

If points be taken along one of the arms of an angle farther and farther from the vertex, their distances from the other arm will at length be greater than any given straight line.

then a certain number of these angles AOC, COR, &c. will be together a right AOK.

From OA cut off that same number of parts OG, GH,... NP, each equal to the given straight line.

Draw PQ to OC,

PQ shall be OG.

Let the A OPQ be turned about OQ into the position ORQ, then about OR and so on, till the number of these As is equal to the number of parts in OP:

thus the ZOA is > the right ▲ KOA;

also PQRZ is

C. G.

OP, for PQRK is > PK, and..> OP;

.. PQ is OG.

3

PARALLEL STRAIGHT LINES.

AXIOM. If one straight line be drawn in the same plane as another, it cannot first recede from and then approach to the other, neither can it first approach to and then recede from the other on the same side of it.

INTRODUCTORY LEMMA.

Through a given point without a given straight line one and only one straight line can be drawn in the same plane with the former which shall never meet it.

Also all the points in each of these straight lines are equidistant from the other.

Then the figure may be turned about AQ till QC falls along QB and AS along AR.

Hence, if there are points in RS on one side of A pearer to BC than A is, so also are there corresponding points on the other side of A nearer to BC than A is; which is impossible.