646. COR. 2.-Lines joining the middle points of the opposite sides of any trapezium bisect each other (?).

647. If two straight lines join the alternate ends of two parallels, the line joining their centres is half the difference of the parallels.

SUG's. We are to prove that EF = † (CD – AB). CHEF = { (CD — AB).

A

B

H

FIG. 363.

648. In any right-angled triangle the line drawn from the right angle to the middle of the hypotenuse is equal to one-half the hypotenuse.

649. The perpendiculars which bisect the three sides of a triangle meet in a common point.

SUG's. First show that the intersection of two of the perpendiculars is equally distant from the three vertices of the triangle. Then that a line drawn from this point to the middle of the third side is perpendicular to it.

650. The three perpendiculars drawn from the angles of a triangle upon the opposite sides intersect

652. If from a point without a circle two secants be drawn, making equal angles with a third secant passing through the same point and the centre of the circle, the intercepted chords of the first two are equal.

SUG. -Prove by revolving one part of the figure.

653. The sum of the alternate angles of any hexagon inscribed in a circle is four right angles.

M

FIG. 365.

pendicular to each other.

654. If two circles intersect in A and B, and from P, any point in one circumference, the chords PA and PB be drawn to cut the other in C and D, CD is parallel to a tangent at P.

655. If two lines intersect, two lines which bisect the opposite angles are per

656. The angle included by two lines drawn from a point within a triangle to the vertices of two of the angles, is greater than the third angle.

SUG's.--The demonstration may be founded on (219) or (231).

657. In a triangle whose angles are 90°, 60°, and 30°, show that the longest side is twice the shortest.

658. Lines which bisect the adjacent angles of a parallelogram are mutually perpendicular.

659. If from any point in the base of an isosceles triangle lines are drawn parallel to the sides, a parallelogram is formed whose perimeter is constant and equal to the sum of the two equal sides of the triangle.

FIG. 366.

FIG. 367.

660. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides of the triangle, their sum is constant and equal to the perpendicular from one of the equal angles of the triangle upon the opposite side.

661. If from any point within an equilateral triangle, three perpendiculars be let fall upon the sides, their sum is constant and equal to the altitude of the triangle.

662. If from a fixed point without a circle two tangents be drawn terminating in the circumference, the triangle formed by them and any tangent to the included arc has a constant perimeter equal to the sum of the first two tangents.

X663. The sum of two opposite sides of a quadrilateral circumscribed about a circle, is equal to the sum of the other two.

X664. If two opposite angles of a quadrilateral are supplemen

tary, it may be circumscribed by a circumference.

665. The square described on the sum of two lines is equivalent to the sum of the squares on the lines, plus twice the rectangle of the lines.

SUG's. Be careful to give the construction fully, and show that the parts are rectangles, etc.

666. The square described on the difference of two lines is equivalent to the sum of the squares on the lines, minus twice the rectangle of the lines.

667. The rectangle of the sum and difference of two lines is equivalent to the difference of the squares described on the lines.

SCH.-The three preceding propositions are but geometrical conceptions and demonstrations of the algebraic formula, (a + b)2 = a2 + 2ab + b2, (a — b)2 = a1 2ab + b2, and (a + b) (a — b) = a2 — b3.

-

a-d

FIG. 371.

To

VARIOUS DEMONSTRATIONS OF THE PYTHAGOREAN PROPOSITION. 668. The square described on the hypotenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.

D.

H

FIG. 372.

1st METHOD.-Let ABC be the given triangle, and ACED the square described on the hypotenuse. Complete the construction, Show that the four triangles are equal. The square HF is (AB BC)'. The student can complete the demonstration.

We will now give a few other figures by means of which the demonstration can be effected, and leave the student to his own resources in effecting it.

9th METHOD.-The truth of the theorem appears also as a direct consequence of (360).

DC

FIG. 377.

669. In an oblique angled triangle the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides diminished by twice the rectangle of the base, and the distance from the acute angle to the foot of the perpendicular let fall upon the base from the angle opposite.

=

SUG's. It is to be shown that AB2 BC2+ AC2 - 2AC x DC. Observe that AD (ACDC)2= AC2 + DC-2AC x DC. Whence, by a simple application of the preceding theorem, the truth of this becomes apparent.

B

D

FIG. 378.

670. In an obtuse angled triangle the square of the side opposite the obtuse angle is equivalent to the sum of the squares on the other two sides, increased by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicula let fall from the angle opposite upon the base produced.

SUG.-The demonstration is analogous to the preceding, C being made obtuse in this case; whence AD = AC + DC, etc.