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PROPOSITION XXXVIII. PROBLEM. 238. To describe a circumference through three points not in the same straight line.
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).
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.
CASE 1. — When the given point is on the circumference.
point on the circumference.
From the centre 0, draw the radius OC.
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.
§ 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.
Let ABC be the given triangle.
Draw the line A E, bisecting 2 A,
Draw E HI to the line A C.
The O KH M is the required.
and EM I to BC.
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).
.. 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.
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.
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).
.:. LAKB= L A B E,
..LAKB= 2 M.
Q. E. F.
PROPOSITION XLII. PROBLEM. 244. To find the ratio of two commensurable straight lines.
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.
Suppose three times with a remainder F D.
Suppose once with a remainder H B.
Suppose once with a remainder K D.
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;
"CD – 18 K D'
.. the ratio of A B _ 41
•• the ratio of CD18
Q. E. F.