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the angle contained by the base and the diameter drawn from the extremity of the base.

33. If a circle be inscribed in a right-angled triangle, the difference between the two sides containing the right angle and the hypotenuse, is equal to the diameter of the circle.

34. If a semicircle be inscribed in a right-angled triangle, so as to touch the hypotenuse and perpendicular, and from the extremity of its diameter a line be drawn through the point of contact to meet the perpendicular produced ; the part produced will be equal to the perpendicular.

GEOMETRICAL ANALYSIS. In the method of analysis we assume the proposition advanced, and then proceed to trace the consequences which follow from this assumption, till we arrive at some known or admitted relation. The reverse of this process constitutes synthesis, or composition, which is the method employed in the preceding pages. In the solution of geometrical problems of more than ordinary difficulty, it is necessary that we should adopt the method of analysis, in order to discover the different steps which must be pursued in the construction. Analysis, observes an eminent geometer, presents the medium of invention ; while synthesis naturally directs the course of instruction.

The following problem is given as an illustration of this method.

D

PROBLEM. Given the hypotenuse of a right-angled triangle,

and the sum of the base and perpendicular, to construct the triangle.

ANALYSIS. Suppose the thing to be done. Let ABC be the triangle required, having bc equal to the given hypotenuse, the angle BAC equal to a right angle, and AB, AC together equal to the given sum of the base and perpendicular. On BA produced take AD equal to Ac, and join cd. Then BD is equal to the sum of

And since ADC is a right-angled isosceles triangle,

A

B

C

AB and

A C.

therefore the angles ADC and ACD are each of them half a right angle. Hence we have the following composition:

SYNTHESIS. Make BD equal to the sum of A B and AC; from D draw (E. 1. 11. and 9.) DC making the angle BDC equal to half a right angle; from B, the other extremity of BD, draw BC equal to the given hypotenuse, meeting dc in c; and from c (E. I. 12.) let fall the perpendicular CA ; then ABC is the triangle required.

For the right-angled triangle Adc is evidently isosceles, and therefore ac is equal to AD, and the sum of AB and AC is equal to BD; wherefore, &c.

THE END.

LONDON :
SPOTTISWOopes and SHAW,

New-street-Square.

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