GEOMETRY 777

From the first of these forms

S>(^-^)and from the second

S<aHence the series is convergent.(At this point the student should solve ten to fifteen exercises

illustrating the preceding theory.)

OEOMETRY.A High School Boy’s Proof of the Pythagorean Theorem.

Theorem: The square on the hypothenuse of a right triangle is equalto the sum of the squares on the two legs.

Given: Right triangle ABC, in,which A B is the hypothenuse; andthe squares, A B E F, A 0 G H, and B 0 K L, on the three sides.To prove: Square ABE F==square A C

G H+square B C K L.Proof: Draw 0 D perpendicular to A B^

and produce 0 D to M in F E.Draw h, altitude of triangle A C D, upon

A C.The measure of square A 0 G H equals

A C X A C.The measure of triangle ADO equals

A C X h

^^Square ACGH

^ACXAC

^2AC

Triangle ACD ACXh h2

The measure of rectangle A D M F equals A P (or A B) xThe measure of triangle ADO equals A D X D C.

A D,

^ Rectangle ADMF^AB X AD

^2AB

Triangle ADC ADXDC "CD"2

Triangle ABC has the altitude C D drawn on the hypothenuse.Then triangle A B 0 is similar to triangle A 0 D. h and C D arehomologous altitudes on the homologous sides A B and A C.

Then2^2^.h CD

Th Square ACGH^

Rectangle ADMFTriangle ACD Triangle ACD

Therefore Square ACGH=Rectangle ADMF.

778 SCHOOL SCIENCE AND MATHEMATICS

Since 0 D is the altitude of the right triangle ABC, upon thehypothenuse, triangle A C D is similar to 0 B D.

Since squares A 0 G H and. B 0 K L are erected on the homologoussides A 0 and B C of the similar triangles, triangle A 0 D: triangleC B D == square A 0 G H : B C K L.But since triangle A 0 D and triangle C B D have the same altitude,

they are to each other as A D is to D B.. �. square A C G H : square BCKL==AD :BDBut AD : B D = rectangle A D F M : rectangle D B M E.. �. square A C G H : square B C K L == rectangle A D F M :

rectangle D B M E.Therefore’square B C K L === rectangle B D M E, and square A 0

G H + square B C K L == rectangle A D F M 4- rectangle B D M E=== square A B E F. Q. E. D.

[Math. Editor’s Note.�This demonstration is original with Mr.Davidson according to his teacher. We take it from the Nautilus.]

A NOTE.

An Extension of the Pythagorean Theorem.

BY HENRY A. CONVERSE, PH.D.,

Davis and Elkins College, ElJdns, TT. Va.

Given three mutually perpendicular axes, OX, OY, and OZ. Let OX = x,OT == y, OZ == s. Then the sides of the triangle XYZ are respectively

]/^y^ l/^+^^ ]/^4-,t:2.Hence the area of XYZ is

s= /l/^+^+l/y^^+lA2-^2 » l/.y24^+l/^24^~t/^T^2"M 2 2

^/^j^2-{/j^p^ 4-1/^4^2 , -\/^2+y2+}/y2+z2+\/22+^22 2

This reduces to

^W^HS)’^\(^\\w

^-(^H?)2. /y^\2. /zx\2

Hence the theorem:

Theorem^�In a tetrahed/ron one of whose trihedral angles is a tri-rectangular trihedral angle, the square of the area of the face oppositethe tri-rectangular trihedral angle is equal to the sum of the squaresof the areas of the other three faces.