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PROPOSITION XXXVI. PROBLEM. 236. Two sides of a triangle and the angle opposite one of them being given, to construct the triangle.
CASE I. When the given angle is acute, and the side opposite to it is less than the other given side.
Let c be the longer and a the shorter given side, and
ZA the given angle. : It is required to construct a A having two sides equal to a and c respectively, and the opposite a equal to given < A. Construct Z D A E equal to the given Z A.
On A D take A B = c.
Draw B C and BC“. Then both the A ABC" and A B ("" fulfil the conditions, and hence we have two constructions.
When the given side a is exactly equal to the I BC, there will be but one construction, namely, the right triangle A B C.
When the given side a is less than B C, the arc described from B will not intersect A E, and her.ce the problem is impossible.
CASE II. When the given angle is acute, right, or obtuse, and the side opposite to it is greater than the other given side.
When the given angle is obtuse.
Take A B equal to a. From B as a centre, with a radius equal to c, describe an arc cutting E A at C, and E A produced at C'.
Join B C and B C'. Then the A A B C is the A required, and there is only one construction ; for the A A B C' will not contain the given < S.
When the given angle is acute, as angle B A C'. There is only one construction, namely, the A B A C (Fig. 1).
When the given Z is a right angle.
There are two constructions, the equal A B A C and BAC" (Fig. 2).
Q. E. F. The problem is impossible when the given angle is right or obtuse, if the given side opposite the angle be less than the other given side,
§ 117 PROPOSITION XXXVII. PROBLEM.
237. Two sides and an included angle of a parallelogram being given, to construct the parallelogram.
Let m and o be the two sides, and C the included
It is required to construct a o having two adjacent sides equal to m and o respectively, and their included 2 equal to Z C.
Draw A B equal to o.
making the 2 A equal to Z C.
On A R take A H equal to m. From H as a centre, with a radius equal to o, describe an arc.
From B as a centre, with a radius equal to m,
Draw E H and E B.
§ 136 (a quadrilateral, which has its opposite sides equal, is a ).
Q. Ę. F.
PROPOSITION XXXVIII. PROBLEM. 238. To describe a circumference through three points not in the same straight line.
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 O as a centre, with a radius equal to 0 A, describe a circle.
O A B C is the O required. For, the point 0, being in the I E O 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.
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, o 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 OH, E M and EH.
ZOM E is a rt. L,
(being inscribed in a semicircle).
Q. E. F. 241. COROLLARY. Two tangents drawn from the same point to a circle are equal.