PROPOSITION XXXVIII. PROBLEM. 238. To describe a circumference through three points not in the same straight line.

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Let the three points be A, B, and C.
It is required to describe a circumference through the three
points A, B, and C.

Draw A B and BC.

Bisect A B and BC. At the points of bisection, E and F, erect is intersecting at 0.

From ( as a centre, with a radius equal to 0 A, describe a circle.

O ABC is the O required. For, the point o, being in the I EO erected at the middle of the line A B, is at equal distances from A and B;

and also, being in the I FO erected at the middle of the line C B, is at equal distances from B and C,

§ 58 (every point in the I erected at the middle of a straight line is at equal

distances from the extremities of that line).
.. the point 0 is at equal distances from A, B, and C,

and a O described from ( as a centre, with a radius equal to 0 A, will pass through the points A, B, and C.

Q. E. F. · 239. SCHOLIUM. The same construction serves to describe a circumference which shall pass through the three vertices of a triangle, that is, to circumscribe a circle about a given triangle.

PROPOSITION XXXIX. PROBLEM. 240. Through a given point to draw a tangent to a given circle.

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Fig. 2.

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Fig. 1.

CASE 1. — When the given point is on the circumference.
Let A B C (Fig. 1) be a given circle, and the given

point on the circumference.
It is required to draw a tangent to the circle at C.

From the centre 0, draw the radius OC.
At the extremity of the radius, C, draw CM I to 0 C.

Then C M is the tangent required, $ 186 (a straight line I to a radius at its extremity is tangent to the O).

CASE 2. — When the given point is without the circumference. Let A B C (Fig. 2) be the given circle, 0 its centre,

E the given point without the circumference.

It is required to draw a tangent to the circle A B C from the point E.

Join () E. On 0 E as a diameter, describe a circumference intersecting the given circumference at the points M and H.

. Draw 0 M and 0 H, E M and E H.
Now
ZOME is a rt. Z,

§ 204
(being inscribed in a semicircle).
... E M is I to O M at the point M;
.. EM is tangent to the O,

§ 186 (a straight line I to a radius at its extremity is tangent to the O). In like manner we may prove H E tangent to the given O.

Q. E. F. 241. COROLLARY. Two tangents drawn from the same point to a circle are equal.

PROPOSITION XL. PROBLEM.
242. To inscribe a circle in a given triangle.

Let ABC be the given triangle.
It is required to inscribe a O in the A A B C.

Draw the line A E, bisecting 2 A,
and draw the line C E, bisecting < C.

Draw E HI to the line A C.
From E, with radius E H, describe the O KM II.

The O KH M is the required.
For, draw EK I to AB,

and EM I to BC.
In the rt. A AK E and A HE
A E= A E,

Iden.
ZEA K= L E A H,

Cons. ..A A KE= A AHE,

$ 110 (Two rt. Á are equal if the hypotenuse and an acute Zof the one be equal respectively to the hypotenuse and an acute 2 of the other).

.. EK= EH,

(being homologous sides of equal A).
In like manner it may be shown EM= EH.

.. EK, E H, and E M are all equal. .. a O described from E as a centre, with a radius equal to E H,

will touch the sides of the A at points H, K, and M, and be inscribed in the A.

§ 174 Q. E. F.

PROPOSITION XLI. PROBLEM. 243. Upon a given straight line, to describe a segment which shall contain a given angle.

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Let A B be the given line, and M the given angle.

It is required to describe a segment upon the line A B, which shall contain < M. At the point B construct Z A B E equal to Z M.

Bisect the line A B at F,

and erect the I FH. From the point B, draw BO I to E B. From 0, the point of intersection of FH and B O, as a centre, with a radius equal to 0 B, describe a circumference.

Now the point 0, being in a I erected at the middle of A B, is at equal distances from A and B,

§ 58 (every point in a I erected at the middle of a straight line is at equal dis

tances from the extremities of that line) ;

.. the circumference will pass through A.
Now
BE is I to 0 B,

Cons. ..BE is tangent to the O,

§ 186 (a straight line I to a radius at its extremity is tangent to the O).

.. L A B E is measured by 1 arc A B, $ 209

(being an Z formed by a tangent and a chord). Also any Z inscribed in the segment A H B, as for instance ZA K B, is measured by ļ arc A B,

(being an inscribed Z).

§ 203

.:. LAKB= L A B E,
(bcing both measured by $ the same arc);

..LAKB= 2 M.
.. segment A H B is the segment required.

Q. E. F.

PROPOSITION XLII. PROBLEM. 244. To find the ratio of two commensurable straight lines.

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Let A B and C D be two straight lines.

It is required to find the greatest common measure of A B and C D, so as to express their ratio in numbers.

Apply C D to A B as many times as possible.

Suppose twice with a remainder E B.
Then apply E B to C D as many times as possible.

Suppose three times with a remainder F D.
Then apply F D to E B as many times as possible.

Suppose once with a remainder H B.
Then apply H B to F D as many times as possible.

Suppose once with a remainder K D.
Then apply K D to H B as many times as possible.

Suppose K D is contained just twice in H B.. The measure of each line, referred to K D as a unit, will then be as follows: —

HB = 2 KD;
FD = H B + KD = 3 KD;
EB = FD + H B = 5 KD;
C D = 3 EB + FD = 18 K D;
A B = 2CD + EB = 41 K D.
.. AB 41 KD

"CD – 18 K D'

.. the ratio of A B _ 41

•• the ratio of CD18

Q. E. F.