.. AEFCAADE (1.8). This suggests the following construction:

Synthesis.-Draw EF || AB (I. 31), then DF is a ;

.. EF = BD (1.34)

=

AD (given), and ▲ A = FEC (I. 29), and DEA = 2C,

:: A ADE = ▲ EFC (I. 26); .. AE = EC.

Q.E.D.

3. Given the middle points of the sides of a triangle, to construct it.

Let A, B, and C be the middle points of the sides. Analysis. Suppose DEF to be the triangle required. It is a well-known proposition (Ex. 100) that the line joining the middle points of two sides of a triangle is to the third

A

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;

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

.. EDF is the ▲ required.

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 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 + EB shall be < lines drawn from A and B to any other point in CD. In going from A to

E

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.

B'F

Synthesis. Draw BF CD, and produce it, making
BF. Join AB' and EB.
B'F, FE is common, and

:: BF

=

=

BFE

= < B'FE,

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′ (I. 20), i.e., >AE + EB,
.. AG + GB > AE + EB.

.. E is the point required.

COR. BED = AEC, for each is equal to 4 FEB'.

..▲ EFC = ▲ ADE (1.8). This suggests the following construction:

Synthesis.-Draw EF || AB (I. 31), then DF is a []; .. EF = BD (I. 34) = AD (given), and 4 AFEC (I. 29), and DEA = 4C,

:: A ADE = ▲ EFC (I. 26); .. AE

=

EC.

Q.E.D.

3. Given the middle points of the sides of a triangle, to construct it.

Let A, B, and C be the middle

points of the sides. Analysis. Suppose DEF to be the triangle required. It is a well-known proposition (Ex. 100) that the line joining the middle points of two sides of a triangle is to the third

A

F

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;

.. AD CB

=

=

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

.. EDF is the ▲ required.

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 rate; 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 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 + EB shall be < lines drawn from A and B to any other point in CD. In going from A to

E

ן

D

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 CD, and produce it, making B'F = BF. Join AB' and EB.

:: BF = B ́F, FE is common, and ▲ BFE = ▲ B'FE, .. BE

=

EB′ (I. 4).

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′ (I. 20), i.e., >AE + EB,
.. AG + GB > AE + EB.

E is the point required.

COR.-4 BED = AEC, for each is equal to 4 FEB’.

.. AEFC = ▲ADE (1.8). This suggests the following construction:

Synthesis. Draw EF || AB (I. 31), then DF is a ☐;

=

.. EF BD (I. 34) = AD (given), and LA FEC

:: A ADE = ▲ EFC (I. 26); :. AE

Q.E.D.

3. Given the middle points of the sides of a tri

angle, to construct it.

Let A, B, and C be the middle

points of the sides. Analysis. Suppose DEF to be the triangle required. It is a well-known proposition (Ex. 100) that the line joining the middle points of two sides of a triangle is || to the third

A

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;

.. AD CB

=

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

.. EDF is the ▲ required.

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 te; he wants to get a drink from the stream, and