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will come on to the road again by the other gate. To do this with the least amount of walking possible, where must he strike the stream?

Let A and B be the

two gates, and CD

the stream,
It is required to find

a point in CD, as
E, so that AE

E F
EB shall be <
lines drawn from
A and B to any

other point in CD. In going from A to

E and then to B, the amount of walking will be the same as if B had been situated as far from the stream as at present, but on the other side of it, as at B'. Now the shortest path from A to B' is along the straight line joining them. The point E, where

this line cuts CD, is the point required. Synthesis.-Draw BF 1 CD, and produce it, making

B'F BF. Join AB' and EB. :: BF = B'F, FE is common, and LBFE = LB'FE,

:: BE = EB' (1.4). :. AE + EB AE + EB AB'. To prove

AE + EB < lines drawn from A and B to any other point in CD, as G. Join GB'. As above, GB = GB, :: AG + GB = AG + GB'. But AG + GB' > AB' (1. 20), i.e., >AE + EB, .. AG + GB > AE + EB.

:: E is the point required. CoR.- BED = L AEC, for each is equal to Z FEB'.

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:: AEFC = AADE (1.8). This suggests the follow

ing construction : Synthesis.-Draw EF | AB (I. 31), then DF is a 0); :: EF = BD (1. 34) = AD (given), and LA = _FEC (1. 29), and < DEA LC, .: A ADE - EFC (1. 26); :: AE = EC.

Q.E.D.

3. Given the middle points of the sides of a triangle, to construct it. Let A,B, and be the middle

points of the sides. Analysis.-Suppose DEF to

be the triangle required. It is a well-known proposi

B
tion (Ex. 100) that the
line joining the middle
points of two sides of a

A
triangle is || to the third

side; :: AC is || DF, BC || DE, and AB || EF. Synthesis. The construction suggested is to draw

through A a line | BC, through B a line || AC,

through C a line || AB (I. 31); :: DC, AF, and BE are parallelograms;

CB = AE (I. 34); :. A is the middle point of DE, and similarly for B and C.

:: EDF is the A required.

:: AD

4. A level field is bounded on one side by a stream flowing in a straight course, and on the opposite side by a straight road, at each end of which there is a gate. A person walking along the road comes to one gate; he wants to get a drink from the stream, and will come on to the road again by the other gate. To do this with the least amount of walking possible, where must be strike the stream ?

Let A and B be the

two gates, and CD

the stream,
It is required to find

a point in CD, as
E, so that AE +

E F
EB shall be <

D
lines drawn from
A and B to any

other point in CD. In going from A to

E and then to B, the amount of walking will be the same as if B had been situated as far from the stream as at present, but on the other side of it, as at B'. Now the shortest path from A to B' is along the straight line joining them. The point E, where

this line cuts CD, is the point required. Synthesis.-Draw BF 1 CD, and produce it, making

Β'F' BF. Join AB' and EB. :: BF = B'F, FE is common, and _BFE = LB'FE,

EB' (I. 4). :. AE + EB AE + EB To prove AE + EB < lines drawn from A and B to

any other point in CD, as G. Join GB'. As above, GB' = GB, :: AG + GB = AG + GB'. But AG + GB' > AB' (1. 20), i.e., >AE + EB, .. AG + GB > AE + EB.

:: E is the point required. Cor.-LBED = _ AEC, for each is equal to – FEB'.

.: BE

AB'.

II. On Loci.

1. In a given line to find a point equidistant from two given points.

А Let A and B be the

given points, and

CD the given line. To find a point in CD

equidistant from A

and B. Analysis.If F is

the pt. required,
then FAB is an isosceles A; and if AB is bisected

at E and EF joined, FE is 1 AB (Ex. 16). Synthesis.-Join AB, bisect it at E (1. 10), and draw

EF I AB (I. 11). Join FA, FB.
AE

EB, and FE is com., and L AEF = L BEF, each being a rt, angle,

:: FA = FB (I. 4). Q.E.F. It is also plain that any point in FE, or FE produced, is equidistant from A and B, so that if in the enunciation of the problem the condition "in a given line" had been omitted, any number of points could have been found to satisfy the remaining condition. Such a problem is said to be indeterminate, and arises in this way, that the given conditions are too few to fix definitely, or determine, the size or the position of the thing required. An example of this would be to find a point when only one condition is given; as,

2. To find a point at a given distance from a given point.

Describe a round the given point with the given distance as radius, and any point in the circumference of the circle satisfies the given condition, and no other point in the same plane does so.

3. To find a point at a given distance from a given line.

If two lines are drawn || the given line at the given distance from it, one on each side of it, all points in these lines will satisfy the given condition, and no other point in the same plane will.

4. To find a point at a given distance from a given circle.

Note.—The distance of a point from a circle is measured along a radius of the circle, produced if necessary.

Points satisfying the required condition will be found on circles concentric with the given circle, and having radii equal to the radius of the given circle increased or diminisbed by the given distance.

5. To find a point equally distant from two given intersecting lines. If a point is equi

B
distant from AB
and AC, perpen-

F
diculars from it on
AB and AC are
equal; but when
DF = DG, and
DA is common,


and <s. DFA and
DGA are rt. angles, ADGA can be proved equal to
ADFA (Ex. 64).

L FAD

į DAG; :. AD bisects < A, and any point in AD is equidistant

..

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