parallelogram HV, having lines HL, MP, RS, each parallel to WV and lines HW, KO, LV, each parallel to TY. Join SW and SV.

TWY

=

Demonstration.-It will be seen that we have here triangle half-parallelogram HY, and parallelogram HY is equal to one-third of the whole figure. Again we have triangle SWV equal to one-half parallelogram RV, which is likewise one-third of whole figure HV.

Then by I. 34 the triangle TWV equals triangle SWV which has a base three times as great as the base WY.

Q.E.F. 2. (a) To a given straight line to apply a parallelogram which shall be equal to a given triangle and have one of its angles equal to that of the given triangle.

(a) See Euclid I., proposition 44.

(b) Let KLO be an equilateral triangle. It is required to apply to one side a parallelogram equal in area having one of its angles equal to the given triangle.

(a) See Euclid I., Proposition 44.

(b) Let KLO be an equilateral triangle.

It is required to

apply to one side a parallelogram equal in area, having one of

its angles equal to the given triangle.

Bisect base LO in M (I. 10,) Join KO. Through point M draw MS parallel to KL and equal to it (I. 31.) Through point K draw KS parallel to LM and equal to it (I. 31).

Then KL MS is parallelogram required. Demonstration.-By I. 41 the parallelogram K L M S is double of the triangle KLM. But triangle KLM is half triangle KLO.

M

Therefore parallelogram KLMS is equal to equilateral triangle KLO, and it has an angle KLM which is common to both.

Q.E.F.

3. (a) If a straight line be divided into any two parts the square on the whole line is equal to the squares on two parts, together with twice the rectangle contained by those parts. (Ity be assumed that parallelograms about the diameter of a square are likewise squares.)

(b) Show algebraically that the squares on the two parts are

always greater than twice their rectangle, except when the line is bisected.

(a) See Euclid, Bk. II., proposition 4. (b) Let m represent smaller portion of line and m + n represent larger portion of line m |

Then the squares of the two parts ad twice their rectangle

Then (m)+(m + n)3

and 2m (m + n)

Subtracting

=

m + n

(m2) + (m + n2)

= 2 (m) (m + n)

= m2 + m2 2m n + n2 = 2m2 + 2 m n + n2

= 2m2 + 2m n

Q.E.D.

Therefore the squares on the two parts are always greater than twice their rectangle by n2.

SECTION IV.

1. (a) In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles, to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

(b) If the triangle be isosceles, the square on the side subtending the obtuse angle is equal to twice the rectangle contained by either side, and the straight line made up of that side, and the straight line intercepted as in the proposition.

Let OMN be an isosceles tri

angle having an obtuse angle ONM.

Then square on OM = twice rectangle MN MP.

By II. 12. Square on OM = square on MN + square on NO + twice rectangle MN NP by Hyp. But MN = ON.

PN

M

Therefore square on MO = twice MN2+ twice rectangle MN NP. But by II. 3., rectangle MP MN = rectangle MN NP + square on MN, and twice rectangle MP MN =

twice rectangle MN NP + twice square on MN.

We have

proved that square on MO = twice rectangle MN NP + twice square on MN.

By Ax. 1. Therefore square on MO = twice rectangle

MP MN.

Q.E.D.

2. (a) Equal straight lines in a circle are equally distant from the centre.

(b) AB are two equal chords of a circle at right angles to each other. Shew that they are sides of a square inscribed in the circle.

(a) See Euclid III. Bk. proposition 14. (b) By Hyp. AB = AC and BAC is a right angle. Join BC, bisect BC in O, join AO and produce to D. Join CD and BD. Then AC DB is a square.

O slym

Since by hyp. BAC is a waded. Dar vi right angle, then by III. 31,\*10191547]°

BAC is a semicircle and BC is a diameter. Again, because by hyp. BA = AC. Then angles ABC and ACB are

each right angle, and since

12

line AD bisects BC in O, and O is the centre of the circle, Then AD is also a diameter, and AO = OD.

B. I., 4. Angle CAO = angle OAB. Therefore I., 32, angle CAO BAO are each rectangle. Then because in triangles CAD, BAD we have (by hyp) CA = BA and AD common to both triangles, and the included angle CAD = to included angle BAD. They by (I. 4) the base .CD = base B D. Again by (III. 22) the angle CAB = angle BDC, and CAD is a right angle. Therefore CDB is a right angle. Again (III., 22) angle ABD = angle ACD, and each is an angle in a semicircle, then each is a right angle—(III. 31.) Then we have all the angles right angles in the figure ACDB. Therefore the figure is a parallelogram, and by (I., 34) AC = BD and AB = CD. Therefore the figure is also equilateral. Q.E.F.

(a) If a straight line touch a circle and from the point of contact a straight line be drawn meeting the circle, the angle

which this line makes with the line touching the circle shall be equal to the angles in the alternate segments of the circle.

(b) If a tangent be defined as the limiting position of a secant, show that the tangent to a circle is perpendicular to the radius drawn to the point of contact.

(a) See III. Bk. Euclid, proposition 18. (b) Let XYZ be a circle of which O is the centre. Draw M N, which is a secant of the circle. Draw OV perpendicular to M N and produce it to S.

By (III., 17) draw tangent HSL. Require to disprove that the angle OSL made by OS meeting HSL is a right angle. Since according to the definition given HSL is parallel to MN. Then by I., 29, the exterior angle OVN = interior and

opposite angle OSL. But the angle OVN is a right angle (construction.) Therefore by Ax. 1, the angle OSL is a right angle, that is the angle made by the perpendicular to the tangent of the circle.

Q.E.D.

GUIDE

TO

PUBLIC ELEMENTARY SCHOOLS AND

TRAINING COLLEGES.

1. An elementary school is a school wherein elementary instruction only is given, and where the fee paid by each scholar does not exceed ninepence per week. (Additional payment may be made for instruction in special subjects, if such instruction is not given during the ordinary school hours.)

2. All elementary schools, whether conducted by private persons or by officially-certificated teachers, are liable to inspection by a Government officer, who is empowered to report on them as held in suitable or unsuitable premises, and as efficiently conducted or the reverse.

3. The following table shows the degree of attainment required of children in such schools. Generally speaking, it is expected that the child shall be able to take the work under ,'Standard I." at seven years of age, that of the following standard at eight, and so on. The part printed in italics indicates work which may or may not be required, according to the character and capabilities of the school. The work is almost

identical in England and Scotland.