D

A

From any point C draw CD equal to B, and CE equal to A (I. 2, 3). From D draw D F at right angles to CD (I. 11). From the point C at the distance C E draw a circle cutting DF at G. Join C G. Then the square described on D G (I. 46) shall be the square required.

Because C D G is a right angle, the square described on C G is equal to the squares described on CD and D G. Therefore the square described on D G alone is equal to the difference between the squares described on CG and CD. But CG is equal to C E, that is, to A; and CD is equal to B. Therefore the square described on D G is equal to the difference between the squares described on A and B.

* (13) Every straight line drawn from the vertex of a triangle to the base is less than the greater of the two sides.

* (14) In the triangle A B C, A D bisecting the angle BAC meets B C in D, and D E, D F parallel to A C, A B respectively meet A B, A C in E, F. Show that D E, D F are equal.

* (15) Prove that any line terminated by two opposite sides of a parallelogram and passing through the intersection of the diagonals is bisected in that point.

* (16) Describe a parallelogram equal to a giyen square, and having an angle equal to half a right angle.

* (17) Describe a square on a given straight line as a dia

meter.

Let A B be the given straight line. from C draw C D at right angles to A B

Bisect it in C (I. 10), (I. 11), produce D C to

E, make CD, CE each equal to A C or C B (I. 3). Join AD D B, BF, EA. Then A E B D shall be the square required.

B

We leave the demonstration of this problem as an exercise for the pupil.

* (18) The line drawn from the right angle of a triangle to the middle of the hypothenuse is equal to half the hypothenuse. We here give the pupil some hints to assist him in the demonstration.

Let D be the middle point of the hypothenuse.

Draw DE, DF parallel to CA, BA. Then in triangles DBE, C D F prove that B E is equal to D F (I. 26, 29), and therefore to E A.

B

F

Then in triangles D B E, DA E, since D E is at right angles to A B (I. 46 Cor.), two sides and included angle are equal each to each, therefore D A is equal to D B.

(19) Any line which is perpendicular to one of two parallel straight lines is also perpendicular to the other.

(20) Any two exterior angles of a triangle are together greater than two right angles.

(21) Draw a triangle which shall be equal to two other giver triangles.

(22) A line drawn from the vertex of an isosceles triangle to the middle point of the base is at right angles to it.

(23) A line making equal angles with the sides of an isosceles triangle will be parallel to the base of the triangle.

(24) Prove that a triangle cannot have two obtuse angles.

*(25) A B is parallel to C D, A D is bisected in E, show that any other straight line drawn through E to meet the two lines will be bisected in that point.

* (26) Prove that the sum of the distances of any point from the angular points of a quadrilateral is greater than half the perimeter of the quadrilateral.

*(27) If the straight line bisecting an exterior angle of a triangle be parallel to a side, the triangle is isosceles.

*(28) If the base of an isosceles triangle be produced both ways, the two exterior angles are greater than two right angles by the vertical angle.

(29) A line drawn through the middle points of the sides of a triangle is parallel to the base.

Let A B C be a triangle, and let the line DE be drawn through the middle points of the sides; then D E is parallel to the base B C.

A

E F

B

Produce D E, and make E F equal to D E.

Join F C. Then in the triangles A D E, EFC; since ED is equal to E F (constr.), and AE is equal to EC (hyp.), and the angle AED is equal to the angle F EC (I. 15); therefore the base F C is equal to the base AD (I. 4), that is to D B. Also the angle A D E is equal to the angle E F C (I. 4), therefore A B is parallel to FC (I. 27), and D B is both equal and parallel to F C.

Therefore D F is parallel to B C (I. 33).

Q. E. D.

(30) Divide a straight line into two parts, so that one of the parts shall be three times the length of the other part.

*(31) If the straight line bisecting the angles at the base of an isosceles triangle be produced to meet, show that they will contain an angle equal to an exterior angle at the base of the triangle.

* (32) Divide a right-angled triangle into two isosceles triangles.

(33) Trisect a right angle.

Let A B C be a right angle. In BC take any point D. On BD describe an equilateral triangle B E D. Bisect the angle BED by the straight line BF; then the angle A B C is trisected by the lines B E, B F.

We leave the demonstration as an exercise for the pupil.

Note.-An exercise may often be worked out in more than one way. Fer example, No. 29 may be demonstrated thus, (The pupil can draw the figure for himself) :—

Let ABC be a triangle; let D, E, be the middle points of A B and A C. Join D E, D C, E B.

Then the triangle D E B is equal to the triangle A E D (I. 38), and the triangle EDC is also equal to the triangle A ED (I. 38); therefore the triangle DEB is equal to the triangle EDC.

But the equal triangles DE B, EDC, are upon the same base D E, therefore they are between the same parallels (I. 39); therefore D E is parallel to B C.

MENSURATION.

Mensuration (Lat. mensura, a measure) is the art of measur ing the dimensions of bodies and figures.

The following tables are used in the measurement of surfaces.

A mile contains 1,760 yards, or 5,280 feet.

The following table of linear measure is used in the measurement of land.

An acre contains 10 square chains, or 100,000 square links.

I. To find the Area of a Square, Rectangle, or other Parallelogram.

Rule.- Multiply the base by the perpendicular height.

Proof. CASE 1.-Let A B C D be a rectangle, of which the