Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION VIII. THEOREM. 286. Two triangles which have their sides respectively perpendicular to each other are similar.

B

[ocr errors]

In the triangles EFD and B A C, let E F, FD and ED,

be perpendicular respectively to A C, BC and A B. We are to prove A EFD and B A C similar.

Place the A EFD so that its vertex E will fall on A B, and the side EF, I to A C, will cut A Cat F'.

Draw F'D' Il to F D, and prolong it to meet B C at H. In the quadrilateral B E D'H, s. E and H are rt. Is. i. ZB+ZED' H=2 rt. Is..

$ 158 But ZE D'F' +ZE D' H=2 rt. .. § 34 ..ZED F=B.

Ax. 3. Now 2C+ZHFC=rt. L,

§ 103 (in a rt. A the sum of the two acute s = a rt. 2); and ZEF D' + 2 H FC=rt. Z. ..LE FD=LC.

Ax. 3. .. A EF D' and B A C are similar. § 280 But AEF D is similar to A EF' D'. $ 279 .. A E F D and B A C are similar.

Q. E. D. 287. SCHOLIUM. When two triangles have their sides respectively parallel or perpendicular, the parallel sides, or the perpendicular sides, are homologous.

Ax. 9. PROPOSITION IX. THEOREM. 288. Lines drawn through the vertex of a triangle divide proportionally the base and its parallel.

[ocr errors][merged small]

In the triangle A B C let HL be parallel to AC, and

let B S and BT be lines drawn through its ver-
tex to the base.
We are to prove

AS ST TO

HO = OR = RL

A BH O and B A S are similar, $ 279 (two A which are mutually equiangular are similar). A BOR and BST are similar,

$ 279 A BRL and B T C are similar, § 279 . AS (SB) ST (BT) TC , HO=lo

. § 278 OR - B

RRI' (homologous sides of similar are proportional).

Q. E. D.

Ex. Show that, if three or more non-parallel straight lines divide two parallels proportionally, they pass through a common point,

PROPOSITION X. THEOREM. 289. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse :

1. It divides the triangle into two right triangles which are similar to the whole triangle, and also to each other.

II. The perpendicular is a mean proportional between the segments of the hypotenuse.

III. Each side of the right triangle is a mean proportional between the hypotenuse and its adjacent segment.

IV. The squares on the two sides of the right triangle have the same ratio as the adjacent segments of the hypotenuse.

V. The square on the hypotenuse has the same ratio to the square on either side as the hypotenuse has to the segment adjacent to that side.

В

A F
In the right triangle A B C, let B F be drawn from the

vertex of the right angle B, perpendicular to the
hypotenuse AC.
I. We are to prove

the A AB F, A B C, and F B C similar.
In the rt. A BA F and BAC,

the acute Z A is common.
.:. the A are similar,

$ 281 (two rt. A are similar when an acute 2 of the one is equal to an acute 2

of the other).
In the rt. A B C F and BCA,

the acute 2 C is common.
.. the A are similar.

§ 281 Now as the rt. A ABF and C BF are both similar to A B C, by reason of the equality of their Ls,

they are similar to each other.

II. We are to prove AF : BF :: BF : FC.
In the similar a AB F and C BF,

A F, the shortest side of the one,
: BF, the shortest side of the other,
:: BF, the medium side of the one,

: FC, the medium side of the other.
III. We are to prove AC : A B :: AB: A F.
In the similar a ABC and A BF,

A C, the longest side of the one,
: A B, the longest side of the other,
:: A B, the shortest side of the one,

: A F, the shortest side of the other. Also in the similar Á A B C and FBC,

A C, the longest side of the one,
: BC, the longest side of the other,
:: BC, the medium side of the one,
: FC, the medium side of the other.

AB AF
IV. We are to prove

B C2 - FC
In the proportion AC : A B :: A B : AF,
A B= AC XAF,

$ 259 (the product of the extremes is equal to the product of the means). and in the proportion AC : BC :: BC : FC, BC = AC XFC.

§ 259 Dividing the one by the other,

A B A C XAF

B C2 ACX FC
Cancel the common factor A C, and we have

AB AF
Ꭴ2 ° F

C2 AC
V. We are to prove

[ocr errors]

(Case III.)

AC = ACX AC.

A B = AC X AF, Divide one equation by the other;

AC AC XAC = then

Ā BAC XAF AI

Q. E. D.

PROPOSITION XI. THEOREM. 290. If two chords intersect each other in a circle, their segments are reciprocally proportional.

Let the two chords A B and EF intersect at the

point 0.
We are to prove AO: E :: 0F : 0 B.

Draw A F and E B.
In the A A O F and EOB,
ZF= 2B,

$ 203
(each being measured by ļ arc A E).
ZA= L E

§ 203 (each being measured by } arc FB). .. the A are similar.

§ 280 (two A are similar when two & of the one are equal to two É of the other). Whence A 0, the medium side of the one, $ 278

: E 0, the medium side of the other,
:: 0 F, the shortest side of the one,
: 0 B, the shortest side of the other.

Q. E. D.

« ΠροηγούμενηΣυνέχεια »