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Proof. 1. Let 1,

diculars fro 2. Then by s bering that the above f

Why?

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3.

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Theorem 13. If a circumference intersects the sides, a, b, c, of a triangle ABC, in the points A1 and A2, B1 and B2, C1 and C2, respectively, then

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theorem, is not a proposition in concurrence or collinearity. It is introduced as necessary to the proof of the very celebrated theorem following.

EXERCISES. 535. By means of Ceva's theorem, prove that the three medians of a triangle are concurrent.

536. Also, that the bisectors of the three interior angles of a triangle are concurrent.

537. Also, that the bisectors of two exterior and of the other interior angles of a triangle are concurrent.

538. Also, that the perpendiculars from the vertices of a triangle to the opposite side are concurrent.

539. By means of Menelaus's theorem, prove that the points in which the three bisectors of the exterior angles of a triangle meet the opposite sides are collinear.

540. Also, that the points in which the two bisectors of two interior angles of a triangle and the other exterior angle meet the opposite sides are collinear.

541. Prove Carnot's theorem for the case in which A is within the circle. (One of several modifications of the above figure.)

542. Investigate Carnot's theorem for a circumscribed triangle; also for an inscribed triangle.

543. The diameters of the two wheels of a bicycle are respectively 1 meter and 0.8 meter; how many revolutions are made by each wheel in going 1 kilometer?

544. If a meter is 0.0000001 of a quarter of the earth's circumference, what is the earth's radius ?

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Theorem 14. If the opposite sides of an agon intersect, they determine three coll

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E

B

Given an inscribed hexagon, ABCDEF, su
DE meet at P, CD and AF at Q, BC and FE
To prove
that P, Q, R are collinear.
Proof. 1. Call the A determined by AB, CD,

as in the figure.

2. Then from Menelaus's theorem,

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6. .. by Menelaus's theorem, P, Q, R ar

NOTE. This theorem, commonly called the Mystic discovered by Pascal at the age of 16.

EXERCISES.

545. To divide a given line AB into two parts, the difference of whose squares shall be equal to the square of a given line CD.

546. To inscribe a square in a triangle.

547. XX' is perpendicular to the diameter AB of a circle at D; a line from A cuts the circumference at C and XX' at P.

Prove that AP AC is

constant. (D may be within or without the circle, etc.)

548. The orthocenter, O, of ▲ ABC is determined by the perpendiculars AD, BE. Prove that AO OD BO⚫ OE.

549. Draw a circle with a central right angle AOB. A and B being on the circumference; bisect AOB by OM, meeting AB at M; draw MPOA; then see if the following is true in general: AB = chord AB+ PA. (Consider special cases, AB = 120°, 180°, 360°.)

550. Given the base and the vertical angle of a triangle; construct it so that its area shall be a maximum.

551. Given the base, the vertical angle, and the rectangle of the sides, to construct the triangle.

552. AB is a diameter of a circle of center O; from any point P on the circumference, PC is drawn perpendicular to AB; from C a perpendicular CE is drawn to OP. Prove that PC is a mean proportional

between OA and PE.

553. On side a of ▲ ABC, point P is taken such that PAC = B. Prove that CP: CB AP2 : AB2. Investigate for three cases, A<, => B.

554. ABC is a triangle right-angled at C; CD L c. Prove that AD: DB CA2: BC2.

=

555. If O, O' are the centers of two fixed circles, such that the circumference of O' passes through O, and if a tangent to circumference of O at T cuts circumference of O' at X, Y, then OX OY is constant. (If the center-line meets the circumference of O' at A, then XTOAYO.)

.....

556. If from any point perpendiculars P1, P2, are let fall on the sides of a regular n-gon circumscribed about a circle of radius r, 2p = nr. 557. If a regular n-gon is circumscribed about a circle of radius r, and perpendiculars P1, P2, Pn, are let fall from the points of tangency of the sides on any tangent, 2p = nr.

.....

558. If O is the orthocenter of triangle ABC, and A', B', C' are the mid-points of a, b, c; Ma, M, Mc are the mid-points of AO, BO, CO; Pa, Pb, Pc are the feet of the perpendiculars from A, B, C to a, b, c ; prove that A', B', C', Ma, Mb, Mc, Pa, Pb, Pc are concyclic. (The "Nine Points Circle.")

SOLID GEOMETRY.

BOOK VI.-LINES AND PLANES IN SPACE.

Section 1.-The Position of a Plane in Space. The Straight Line as the Intersection of two Planes.

DEFINITIONS. Through three points, not in a straight line, any number of surfaces may be imagined to pass.

For example, through the points A, B, C the surfaces P and S may be imagined to pass.

A surface is called a plane surface if it possesses the following

quality Through three points

B

P

one plane surface, and only one, can pass, if the three points are not in the same straight line.

These definitions are repeated from the Plane Geometry.

Solid Geometry treats of figures whose parts are not all in one plane.

POSTULATE 7. From the definition of a plane, the following postulate may be assumed and added to those already stated: If two points of a straight line lie in a plane, the whole line lies in that plane.

EXERCISE. 559. Show that if there are given four points in space, no three being collinear, the number of distinct straight lines determined by them is six; if there are five points, the number of lines is ten.

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