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PROPOSITION XXIII. PROBLEM 236. To construct a triangle, when two sides and an angle opposite one of them is given.

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Given. Lines a and b, and 2 A.

Required. A triangle, two sides of which are equal to a and b, a being opposite an angle equal to A.

Construction. Draw Z GEI= L A.
On EG, lay off ED=b.

From D as a center with a radius equal to a, draw an arc intersecting EI in F and F".

Both A EDF and EDF fulfil the required conditions.

DISCUSSION. If the arc intersects the base twice, there will be two solutions, and if it touches the line, but one. If it does not touch the line, a solution is impossible.

Ex. 400. In the Prop. XXIII, how many solutions are possible, when angle A is obtuse ? right ? acute ?

Ex. 401. Construct a right triangle, having given the hypotenuse and

one arm.

Ex. 402. Is it possible to solve the preceding problem by drawing the hypotenuse first ?

237. A triangle may be constructed if the following parts are given:

(1) Three sides.
(2) Two sides and the included angle.
(3) Two angles and the included side.

(4) Two angles and a non-included side.

(5) Two sides and the angle opposite one of them. To construct a triangle, three independent parts must be given.

Ex. 403. Are the three angles of a triangle three independent parts, and can a triangle be constructed when the three angles are given ?

PROPOSITION XXIV. PROBLEM 238. From a given point, to draw a tangent to a given circle.

A

I. When the given point, A, is in the circumference.

Hint. — What is the angle formed by a radius and a tangent at its extremity ?

II. When the given point, A, is without the circle.

B В

Construction. Join A, and O the center of the given circle.

On OA as a diameter, construct a circumference, intersecting the given circumference in B and C.

Then AC and AB are the required tangents.
Hint. — Show that É ACO and OBA are right angles.

Ex. 404. Construct a line tangent to a given circle and parallel to a given line.

Ex. 405. Construct a line tangent to a given circle and perpendicular to a given line.

PROPOSITION XXV. PROBLEM 239. To inscribe a circle in a given triangle.

B

D Given. A ABC. Required. To inscribe a circle in A ABC. Construction. Bisect the B and C. From 0, the intersection of the bisectors, draw ODI BC.

From O as a center, with a radius equal to OD, draw circle, which is the required one.

[The proof is left to the student.]

240. DEF. A circle touching one side of a triangle and the prolongations of the other two sides is an escribed circle.

a

Ex. 406. Construct the three escribed circles of a triangle.

Ex. 407. The bisector of an angle of a triangle ineets the circumference of the circumscribed circle in a point which is equidistant from the other two vertices of the triangle and the center of the inscribed circle.

PROPOSITION XXVI. PROBLEM 241. Upon a given straight line, to construct a segment of a circle which shall contain a given angle.

[blocks in formation]

Given. Line AB and Z M.

Required. To construct a segment of a circle on AB which shall contain 2 M.

Construction. At A, draw ZBAC = Z M.
At A, draw AEI to AC.
Draw the perpendicular-bisector of AB, intersecting AE at 0.

From O as a center, with a radius equal to 0A, draw a circle ADB.

ADB is the required segment.
Proof. AC is a tangent to the circumference. (191)

.. Z CAB is measured by į arc AB. (224) But 2D is measured by arc AB;

(220) .. ZD= 2 CAB ZM.

Q.E.F.

ANALYSIS OF PROBLEMS 242. An analysis of a problem is a course of reasoning by which a construction is discovered. Although no rules can be given which apply to all constructions, the method explained in the following exercises may be used in many problems.

Ex. 408. To construct a triangle having given one side, the corresponding median, and the altitude to another side.

Given. Three lines, b, ha, My.

Required. A triangle having one side equal to b, the corresponding median equal to me, and the altitude upon another side equal to he

B

br

hat

тън

A

E

Analysis. (1) Suppose ABC is the required triangle. (2) Then we know CA(=b), CE and EA

= ha, BE (= mg), and É ADC and ADB (= rt. ).

(The student is advised to mark the known parts as indicated in the diagram, or to draw them in a different color from the other lines.)

(3) Examine all triangles in the figure, and see if one can be constructed. The rt. A ADC can be constructed, having given two sides.

(4) Make this triangle the basis of the construction.
Hence,
Construction. Draw DA=ha

B В
At D, draw FH I to AD.
From A as
a center, with a

D
radius equal to b, draw an arc in-
tersecting DH in C. Draw AC.

Bisect AC in E, and from E as a center, with a radius equal to My, describe an arc, meeting FH in B.

H ABC is the required triangle.

F

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