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 A B C be a triangle, and A D the perpendicular

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

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

Cons.
ZECA is a rt. 2,

§ 204 (being inscribed in a semicircle). ..ZBDA= Z ECA. ZB=LE,

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

..A ABD and A E C are similar, § 281 (two rt. A having an acute Z of the one equal to an acute 2 of the other are

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

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

ВА А )
or,

$ 278 E A – AC ..BAX AC = EA X AD.

Q. E. D.

PROPOSITION XX. THEOREM.

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

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

AC and B D its diagonals.
We are to prove BDXAC = A B XC D+AD X BC.
Construct ZA BE= Z DBC,
and add to each ZEBD.
Then in the A A B D and B C E,
ZAB D= _ C BE,

Ax. 2
and
<BDA = LBCE,

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

.. A AB D and BC E, are similar, $ 280 (two A are similar when two & of the one are equal respectively to two e

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,

or,

§ 278

AD BD

CE = BC'
(the homologous sides of similar A are proportional).

UBDXCE= A D X BC.
Again, in the D A B E and BCD,

LABE= Z DBC,

Cons.

and

$ 203

Z BAE = 2 BDC,
(each being measured by 1 of the arc BC).

..S AB E and B C D are similar, § 280 (two S are similar when two es of the one are equal respectively to two

of the other).

Whence A B, the longest side of the one,

: BD, the longest side of the other,
:: A E, the shortest side of the one,
: CD, the shortest side of the other.

§ 278

АВАЕ

BD - CD'
(the homologous sides of similar A are proportional).

..BD X A E= AB XC D.

Adding these two equalities,

BD (A E + C E) = A B XCD + AD X BC, or BDX AC= A B XCD + ADX BC.

Q. E. D.

Ex. If two circles are tangent internally, show that chords of the greater, drawn from the point of tangency, are divided proportionally by the circumference of the less.

ON CONSTRUCTIONS.
PROPOSITION XXI. PROBLEM.

302. To divide a given straight line into equal parts.

A

Let A B be the given straight line.
It is required to divide A B into equal parts.
From A draw the indefinite line A 0.

Take any convenient length, and apply it to A O as many times as the line A B is to be divided into parts.

From the last point thus found on A 0, as C, draw C B.

Through the several points of division on A O draw lines Il to C B.

These lines divide A B into equal parts, $ 274 (if a series of lls intersecting any two straight lines, intercept equal parts

on one of these lines, they intercept equal parts on the other also).

Q. E. F.

Ex. To draw a common tangent to two given circles.
I. When the common tangent is exterior.
II. When the common tangent is interior.

PROPOSITION XXII. PROBLEM. 303. To divide a given straight line into parts proportional to any number of given lines. н к

в

m

Let A B, m, n, and o be given straight lines.

It is required to divide A B into parts proportional to the given lines m, n, and o.

Draw the indefinite line A X.
On A X take AC=m,

=n,
and . EF=0.
Draw FB. From E and G draw E K and C H |to F B.

K and H are the division points required.
For
(A K\ Α Η Η K KB

§ 275 (a line drawn through two sides of a A ll to the third side divides those

sides proportionally).

.. AH : HK : KB : : AC :CE: EF. Substitute m, n, and o for their equals A C, C E, and E F. Then AH : HK: KB : : m :n : 0.

Q. E. F.