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At A draw AC at right angles to AB (1. 11);
at B make the angle ABD equal to the angle BAD (1. 23); through A, B drawAE, BE parallel to BD,AD respectively (1. 31). Then the figure DAEB shall be the square required.
Proof. Because the angle DBA is equal to the angle DAB (con.),
therefore the side DB is equal to the side DA (1. 6);
and the opposite sides of a parallelogram are equal (1. 34);
Again, the angle EAB is equal to the angle ABD (1. 29),
The parallelogram DAEB has the angle DAE a right angle,
therefore the figure DAEB is a square (def. 30),
and it is described with the given straight line AB as a diameter. Q.E.F.
On a given straight line describe an isosceles triangle having each of the sides equal to a given straight line.
Let AB be the given base, and C the given length of the equal sides.
It is required to describe upon AB an isosceles triangle having each of the sides equal to C.
Make AK, BL each equal to C (1. 3).
From the centre A, at the distance AK, describe the circle DEF (post. 3);
from the centre B, at the distance BL, describe the circle DGF;
and from D, one of the points in which the circles cut one another,
draw the straight lines DA, DB to the points A, B (post. 1). Then ABD shall be an isosceles triangle having its sides DA, DB each equal to C.
Proof. Because A is the centre of the circle DEF, therefore AD is equal to AK (def. 15),
and it is therefore equal to C (ax. 1);
also similarly BD is equal to C.
Wherefore ABD is an isosceles triangle, and each of the sides DA, DB is equal to C. Q.E.F
Through a given point on a side of a square, draw a line to the opposite side which shall bisect the square.
Let ABCD be the given square, and E the given point in the side BC.
It is required to bisect the square ABCD by a straight line drawn through E to the side AD.
From DA cut off DF equal to BE (1. 3), and join EF. Then the straight line EF shall bisect the square ABCD. Draw the diagonal BD, cutting EF in G.
Proof. In the triangles BGE, DGF,
the angle BEG is equal to the angle DFG (1. 29),
and the triangles BGE, DGF have been proved to be equal;
Wherefore the straight line EF bisects the square ABCD.
47. What is the length of a ladder which will just reach from the edge of a ditch 5 yards wide, to the top of a wall on the opposite side 12 yards high?
1. The angles at the base of an isosceles triangle are equal. Show from this that the opposite angles of a rhombus are equal to each other.
Let ABCD be a rhombus.
Then the angle ABC shall be equal to the angle ADC, and the angle BAD shall be equal to the angle BCD.
Proof-Because AB is equal to AD (hyp.),
the angle CBD is equal to the angle CDB;
wherefore the whole angle ABC is equal to the whole angle ADC (ax. 2).
Similarly, by drawing the diagonal AC, it can be proved that the angle BAD is equal to the angle BCD. Q.E.D.
2. Any two sides of a triangle are greater than the third side. Hence prove that the sides of a four-sided figure are together greater than the two diagonals.
Let ABCD be a quadrilateral figure, of which AC, BD are the diagonals.
Then the sides AB, BC, CD, DA shall be together greater than the diagonals AC, BD.
Proof-The sides AB, BC are greater than the side AC
the sides AD, DC are greater than the side AC;
the sides BA, AD are greater than the side BD;
and the sides BC, CD are greater than the side BD; therefore twice the sides AB, BC, CD, DA are together greater than twice the diagonals AC, BD;
wherefore the sides AB, BC, CD, DA are together greater than the diagonals AC, BD. Q.E.D.
3. Through a given point within an angle BAC to draw a straight line cutting off equal parts from AB, AC.
Let D be the given point within the angle BAC.
It is required to draw through D a straight line which shall cut off equal parts from AB, AC.
Produce BA to E (post. 2), and bisect the angle CAE by the straight line AF (1. 9);
through D draw GH parallel to AF, cutting AB, AC in G, H respectively (1. 31).
Then AG shall be equal to AH.
Proof-Because GH is parallel to AF, and EB falls upon them,
the exterior angle EAF is equal to the interior and opposite angle AGH (1. 29);
and because GH is parallel to AF, and AH meets them,
the angle FAH is equal to the alternate angle AHG (1. 29):