121. Prove that the median lines of a triangle divide each other into parts, which are to each other as 1:2.

122. Prove that the line in a trapezium, joining the middle points of the sides which are not parallel, equals half the sum of the parallel sides.

123. The parallel sides of a trapezium are produced in opposite directions, so that each of the prolongations equals the opposite side. Shew that the line, joining the extremities of the prolongations, bisects one of the diagonals of the trapezium, and cuts the other in a point lying equally distant from one extremity of the diagonal as the point of intersection of the diagonals from the other extremity. 124. From a given point 0 to a given line, any line OA is drawn, and on this a point a is marked off, the product OA. Oa being given. Shew that a describes a circle through 0, when A traverses the given line.

125. In a triangle BAC the angle A is a right angle, and CD bisects the angle C. Prove that AB. AD AC.(BC—AC). 126. Two circles touch each other externally, shew that the part of their common tangent between the points of contact is a mean proportional between the diameters. 127. In a triangle ABC the median line AD is drawn, and through A a line AE ‡ BC. Any line through D cuts AE in E, AB in F, and AC in G; prove that

DF: DG EF: EG.

128. Through the point of intersection of the diagonals of a trapezium, a line is drawn parallel to the parallel sides. Prove that these sides are to each other as the parts in which the diagonal divides the not parallel sides. 129. AD is a diameter of a circle and AB, BC, and CD tangents. Shew that both diagonals in ABCD bisect the perpendicular from the point of contact of BC to the diameter.

130. Two given circles touch each other in 0. Through 0 draw any line, cutting the circles in A and B. Shew

that the radius of a circle, touching the one circle in A and passing through B, is constant.

131. A circle touches another internally at A. BC is a chord in the large circle, touching the lesser one at D. Shew that AD bisects the angle BAC.

132. From a point in the circumference of a circle, perpendiculars are drawn to two tangents, and to the chord, joining their points of contact; prove that the last perpendicular is a mean proportional between the two first. 133. In a triangle ABC a line is drawn, cutting AB in c, AC in b, and BC produced in a. Prove that

Draw Cm AB, and employ bm C ~ bc A, maⱭ≈caB, and eliminate Cm from the two equations obtained thus. 134. From the angular points of a triangle ABC, the lines Aa, Bb, and Cc are drawn to the opposite sides and cutting each other in the same point 0; prove that Ba. Ac. Cb Ab. Ca. Bc.

Apply ex. 133 to ▲ ABb, cut by Cc, and to BbC, cut by Aa; thereupon eliminate AC from the two equations. 135. Demonstrate the proposition in 87, by the help of perpendiculars drawn from the extremities of the divided side to the bisecting line.

136. In two given circles with centres C and c draw any two parallel radii CA and ca. Shew that Aa cuts the line of centres in a fixed point (the exterior centre of similitude), when the two radii go in the same direction, and in another fixed point (the interior centre of similitude), when the radii go in opposite directions. What construction of the common tangents to the circles can be deducted from this?

137. The points a and b divide the line AB harmonically; prove that A and B also divide ab harmonically.

When A and B are fixed, and a moves from A to B,

how does b move simultaneously? When AB

k,

138. Shew that OA (Ex. 137) is a mean proportional between Oa and Ob, when O is the middle point of AB.

II. THE RIGHT-ANGLED TRIANGLE.

95. The altitude on the hypothenuse of a right-angled triangle, divides it into two triangles, which both are similar to the whole triangle, and therefore similar to one another; for we have

From this different proportions are deducted, of which

we will mention the most important.

a) the altitude is a mean proportional between the parts of

b) one of the sides containing the right angle is a mean proportional between its projection on the hypothenuse and the whole hypothenuse, and that

c) the product of the sides equals the product of the altitude and the hypothenuse.

The most important proposition of the right-angled triangle is the one discovered by Pythagoras in the 6th century B. C.

d) The square on the hypothenuse equals the sum of the squares on the sides.

This proposition shews that the one side of a rightangled triangle can be calculated, when the two others are known; we find

By the help of the propositions concerning the right-angled triangle, when two of the parts a, b, h, a, ß, and p are known, the others can be found; some of these problems, however, lead to equations of the second degree.

If 20 is obtuse, and the opposite side is c, and the sides containing it a and b, then c> Va2 + b2; if ≤C is acute, then c<Va2+b2; for if the triangle be compared to the rightangled triangle with the sides a and b, then in the first case c>h, in the second c<h. (46).

96. To construct a mean proportional between two given lines, a and b.

a) On the greater line b mark off a (AB), and on bas diameter describe a semicircle; at the extremity of a raise

meet together raise a perpendicular (BD), which then will be the required line (95, a).

.a

х

As == 2, then x = Vab. A line which can be expres

b

=

sed by known lines in this form is therefore constructed in one of the methods indicated; if, for example, x = a√5 V5a.a, then a will be a mean proportional between a and 5a. We could also put x V(2a)2 + a2, and therefore construct a as hypothenuse of a right-angled triangle, in which the one side is a and the other 2a.

2 =

=

Vabcd is constructed by putting ab= x2, cd = y2; x and y are then found by 96; thereupon z is found as hypothenuse of a right-angled triangle with sides x and y.

III. THE POTENSE OF A POINT WITH REGARD TO A CIRCLE.

97. a) When two chords of a circle intersect, the product of the two parts of the one chord equals the product of the parts of the other chord.

Draw the lines Ab and Ba; then

α

B

b) When two secants cut each other, the product of the one secant and the part outside the circle equals the product of the other secant and the part of it outside the circle.

The former demonstration also applies to this case. As the proposition holds good for every position of the secants, it also holds good in the case in which b