(38.) If two equal circles touch each other externally, and through the point of contact another be described. with the same radius; the area contained by the convex circumferences cut off from the touching circles, and the part of the third without them, is equal to the area of the quadrilateral figure formed by lines drawn from the points of intersection to the point of contact, and to the point where the third circle is cut by a tangent drawn to the point of contact of the two circles.

E

Let two equal circles touch each other in A; and through the point of contact let an equal circle ABC be described, cutting the former in B and C. Join AB, AC; and to the point A let a tangent AD be drawn; join BD, DC. The area contained by AEB, AFC and the intercepted arc BDC is equal to the quadrilateral figure ABDC.

Since DA touches the circle AEB, the angle DAB is equal to the angle in the alternate segment, and .. equal to the angle in the segment BCA, i. e. equal to the angle BDA, whence BA= BD; .. the segment BEA is equal to the segment BGD; and AEBGDA is equal to the triangle ABD. In the same manner it may be shewn that AFCHDA is equal to the triangle ACD'; .. AEBGHCFA is equal to the quadrilateral figure ABDC.

(39.) If a straight line be divided into any two parts, and upon the whole and the two parts semicircles be described; and from the point of section a perpendicular

be drawn, on each side of which circles are described touching it and the semicircles; these circles will be equal.

Let AB be divided into any two parts in C; and on AB, AC, CB let semicircles be described; from C draw the perpendicular CD, on each side of which let a circle be described touching

A

E H

G

K

the perpendicular and each of the semicircles. These circles are equal.

Let EFGH touch the perpendicular in H, and the semicircles in F and G. Draw the diameter EH parallel to AB. John FE, EA; AF will be a straight line (ii. 35.). Produce it to meet the perpendicular in D. Join FH, HB; FB will also be a straight line, and perpendicular to AD at the point F. Join EG, GC, HG, GA; EC and HA will also be straight lines. Produce AH to I. Join BI; it will be perpendicular to AI, and pass through D, since the perpendiculars to the three sides of the triangle AHB meet in a point. And since the angles AGC, AIB are equal, EC is parallel to DB, .. AD: DE :: AB : BC;

and AC, HE are parallel,

.. AD : DE :: AC : EH,

.. AB: BC :: AC: EH,

In the same manner it may be proved that
AB AC: BC: KL

KL being the diameter of the other circle drawn parallel

to AB. Hence EH= KL, and the circles are equal.

SECT. IX.

(1.) GIVEN one angle, a side adjacent to it, and the difference of the other two sides; to construct the triangle.

Let CB be equal to the given side; draw the indefinite line CA, making with it an angle equal to the given angle; and cut off CD equal to the given difference. Join

B

BD; and make the angle DBA equal to BDA; ABC is the triangle required.

The angle DBA being equal to ADB, the side AD is equal to AB; and the differences between CA and AB is equal to CD, i. e. to the given difference.

(2.) Given one angle, a side opposite to it, and the difference of the other two sides; to construct the triangle.

In any line CA, (see last Fig.) take CD equal to the given difference; make the angle CDB greater than a right angle by half the given angle; from C draw CB equal to the given side, and meeting DB in B; and make the angle DBA equal to BDA; ABC is the triangle required.

Since the angles ABD, ADB are equal, AB is equal to AD; ... the difference between CA and AB is equal

to CD, i. e. to the given difference. Also the angle at A is equal to the difference between the angles CDB, DBA, or CDB, BDA, i. e. to the given angle. And CB was made equal to the given side.

(3.) Given the base and one of the angles at the base; to construct the triangle, when the side opposite the given angle is equal to half the sum

and a given line.

the other side

B

Let AB be the given base, and ABC the given angle; produce CB to D, making BD equal to the given line. Join AD; and from B to AD draw BE, equal to half BD. From A draw AC parallel to BE; ABC is the triangle required. For AB and ABC are made equal to the given base and angle; and since BE is parallel to AC,

AC: CD: BE: BD: 1 2.

(4.) Given the base of a right-angled triangle, and the sum of the hypothenuse and a straight line, to which the perpendicular has a given ratio; to construct the triangle.

Let AB be equal to the given base. From B draw BC perpenular to it, and such that it may to the given sum, the given

ratio. Join CA, and produce it; and from B to CD, draw BD equal to the given sum.

From A draw AE

perpendicular to AB; ABE is the triangle required. For AE being parallel to BC,

AE: ED :: BC : BD, i. e. in the given ratio; .. AE is equal to the perpendicular; and AB was made equal to the given base.

(5.) Given the perpendicular drawn from the vertical angle to the base, and the difference between each side and the adjacent segment of the base made by the perpendicular; to construct the triangle.

FB

From any point C in an indefinite line AB, erect a perpendicular CD equal to the given perpendicular; and take CE equal to the given difference between the side and AC adjacent segment on the opposite side of the perpendicular. Also take CF equal to the other given difference, Join ED, FD; and make the angle FDA = DFA, and EDB=DEB; ADB is the triangle required.

Since the angles ADF, AFD are equal, AD=AF, ..the difference between AD, AC is equal to CF, i. e. to the given difference. In the same manner the difference between BD and BC is equal to CE, i, e. to the given difference.

(6.) Given the vertical angle, and the base; to construct the triangle, when the line drawn from the vertex cutting the base in any given ratio, bisects the vertical angle.