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PROBLEM E.

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

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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

Join HK:

required.

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.

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

From A let fall AQ to BC.

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

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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.

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Let AOC be the given angle,

then a certain number of these angles AOC, COR, &c.

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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 ▲ 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 4 KOA;

also PQRZ is > OP, for PQRK is > PK, and .'. > OP; .. PQ is > OG.

C. G.

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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.

Let A be the given point

without the given straight R

line BC.

From A draw AQL to B

BC, and through A draw

RAS 1 to AQ.

C

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

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Hence, if there are points in RS on one side of A nearer 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.

And if points in RS on one side of A are farther from BC than A is, so also are there corresponding points on the other side of A farther from BC than A is; which is impossible.

.. all points in RS are equidistant from BC.

Similarly all points in BC are equidistant from RS.

Again, if any line be drawn through A other than RAS, and points taken along that part of it which is on the

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same side of RAS as BC is, farther and farther from A, their distances from RAS will at length be greater than AQ:

(I. G.)

.. the straight line thus drawn will at length cut BC. Hence through A no other straight line besides RAS can be drawn which shall never meet BC.

DEF. Straight lines, lying in the same plane and which will never meet, though produced ever so far both ways, are called parallel. [The name being due to the property established in the foregoing Lemma.]

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