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Steps of the Demonstration to Part 1st,

That the Zino is a right L. 1. Prove that _ EAB = Z ABE; and EAC = _ ACE,

that :: whole BAC = US ABC + ACB, 3.

that Z BAC = _ FAC, and :: each is a rt. Z.

Part 2nd. That < in segment > 1 o is < right Ze is proved by showing that since Z BAC of A ABC is a right 2, .. the Z ARC < right Z.

Steps of the Demonstration to Part 3rd,

That _ in a segment < } o is > right L. 1. Prove that ZS ABC + ADC = 2 right Zs . 2. that < ADC > right 2.

PROPOSITION XXXII. Theorem. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.

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PROPOSITION XXXIII. Problem. Upon a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle.

The proof of case 1st, in which < c is a right L, is merely the showing that _ AHB is also a right 2, because she is do.

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Steps of the Demonstration to Case 2nd,

In which c is not right 2. 1. Prove that (in As AFG, BFG) base AG = base by, that :: a o described from g with distance

GA will pass through B, that AD touches the o, that _ DAB and :: Zc= _ in alternate

segment AEB (or aho in fig. 2.)

PROPOSITION XXXIV.

Problem. From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.

E ------

B---------F

Steps of the Demonstration. 1. Prove that FBC = _ in alternate segment BAC,

that _ in the alternate segment bac = _ D.

PROPOSITION XXXV. Theorem. If two straight lines within a circle cut each other, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

: In case 1st, E is the centre of the o, and :. it is evident that AE X EC = BE X ED.

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Steps of the Demonstration to Case 2nd, (In which BD passes through the centre, and cuts' ac, which does not pass through the centre, at rt. Zs.)

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The last case (in which neither ac nor bo pass through the centre) is proved by showing that

AE X EC and be X ED are each = GE X EH, and :: = each other.

PROPOSITION XXXVI. Theorem. If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it, the rectangle B/. contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

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Steps of the Demonstration to Case 1st, (In which da passes through the centre). 1. Prove that Ebd is a right 2,

that AD X DC + ECS – ED',
that AD X DC + EB' — EBS + BD',
that AD X DC = BD®.

Steps of the Demonstration to Case 2nd, (In which da does not pass through the centre). 1. Prove that AF = FC,

that AD X DC + Fc = FD',

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