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PROPOSITION XVII. THEOREM. 296. The homologous altitudes of two similar triangles have the same ratio as any two homologous sides.

In the two similar triangles ABC and A' B'C', let the altitudes be B O and B'O'.

. Bo AB

We are to prove BO = A' B' *

In the rt. A BO A and B' O' A',
ZA= LA

§ 278 (being homologous ts of the similar D A B C and A' B' C').

.. A B O A and A B' O' A' are similar, § 281 (two rt. A having an acute Z of the one equal to an acute 2 of the other are

similar).

.. their homologous sides give the proportion

BO AB
B'O A' B''

Q. E. D

297. Cor. 1. The homologous altitudes of similar triangles have the same ratio as their homologous bases.

.:. BO

In the similar A A B C and A' B'C',
AC _ A B

$ 278
A' C A' B?
(the homologous sides of similar A are proportional).
And in the similar A BO A and B' O' A',

BO - AB
BO = A' B'

§ 296 А с

Ax. 1 B' 0 = A'C' 298. Cor. 2. The homologous altitudes of similar triangles have the same ratio as their perimeters.

Denote the perimeter of the first by P, and that of the second by P'.

Р А В
Then

§ 295 P = A' B' (the perimeters of two similar polygons have the same ratio as any two

homologous sides).

ВО АВ.
But

§ 296
B' O = A' B'
BO P

Ax. 1 ** B' O =

Ex. 1. If any two straight lines be cut by parallel lines, show that the corresponding segments are proportional.

2. If the four sides of any quadrilateral be bisected, show that the lines joining the points of bisection will form a parallelogram.

3. Two circles intersect; the line A H K B joining their centres A, B, meets them in H, K. On A B is described an equilateral triangle A B C, whose sides B C, A C, intersect the circles in F, E. FE produced meets B A produced in P. Show that as PA is to P K so is C F to CE, and so also is PH to PB. PROPOSITION XVIII. THEOREM. 299. In any triangle the product of two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle together with the square of the bisector.

Let Z BAC of the A A B C be bisected by the straight

line AD.
We are to prove BAX AC = B D X DC + A D.

Describe the O A B C about the A A BC;
produce A D to meet the circumference in E, and draw EC.
Then in the A ABD and A EC,
Z BAD= Z CA E,

Hyp. ZB=LE,

§ 203 (each being measured by the arc A C).

.. A ABD and A E C are similar, § 280 (two A are similar when two of the one are equal respectively to two Ś

of the other).
Whence BA, the longest side of the one,

: EA, the longest side of the other,
: : A D, the shortest side of the one,
: A C, the shortest side of the other;

BA AD
or,

EA - AC'
(homologous sides of similar @ are proportional).
..BA XA C= EA X A D.

E A XAD= (ED + A D) AD,
.:. BA X. A C = E D X A D + A D?.
EDX AD= B D X DC,

$ 290 (the segments of two chords in a which intersect each other are

reciprocally proportional). Substitute in the above equality B D X D C for ED X AD,

then BAX AC = B D X D C + A D.

§ 278

But

But

Q. E. D.

PROPOSITION XIX. THEOREM. 300. In any triangle the product of two sides is equal to the product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle.

Let ABC be a triangle, and A D the perpendicular

from A to BC.
Describe the circumference A B C about the A A BC.

Draw the diameter A E, and draw E C.
We are to prove BA X AC = E A X A D.
In the A ABD and A EC
ZBD A is a rt. 2,

Cons.
ZEC A is a rt. Z,

$ 204 (being inscribed in a semicircle). ..ZB DA = Z ECA. ZB=LE,

§ 203 (each being measured by į the arc A C).

.. A ABD and A E C are similar, § 281 (two rt. A having an acute of the one equal to an acute Z of the other are

similar).
Whence BA, the longest side of the one,

: EA, the longest side of the other,
: : AD, the shortest side of the one,
: A C, the shortest side of the other ;
ВА А)

§ 278
E A – AC
. B A X A C = E 1 X 4 D.

Q. E. D.

PROPOSITION XX. THEOREM. 301. The product of the two diagonals of a quadrilaterai inscribed in a circle is equal to the sum of the products of its opposite sides.

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Let A B C D be any quadrilateral inscribed in a circle,

AC and B D its diagonals.

We are to prove BD XAC = A B XC D + A D X BC.
Construct ZA BE= Z DBC,
and add to each Z EBD.
Then in the A A B D and BC E,
LABD=LCBE,

Ax. 2
and
ZBDA= LBCE,

§ 203 (each being measured by } the arc A B).

.. A A B D and B C E, are similar, $ 280 (two A are similar when two É of the one are equal respectively to two

of the other).

Whence AD, the medium side of the one,

: CE, the medium side of the other,
:: B D, the longest side of the one,
: BC, the longest side of the other,

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