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But (demonstr. of E. 3. 3. and constr.)

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.. 2 AEK+ 2

Take from both the

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CAK2ACK;

ACB = ▲ ACK.

ACB and there remains

Z AEK or KEC=/ BCF.

L

Again (E. 32. 1.) the EGB= ECG+ 2 CEG:

And it has been shewn that the

CEG

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BCF

PROP. XXVI.

35. THEOREM. The perpendiculars let fall from the three angles of any triangle upon the opposite sides, intersect each other in the same point.

Let ABC be a A; the perpendiculars let fall

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from the three A, B, C, on the sides opposite to them, intersect each other in the same point.

For draw (E. 12. 1.) AD 1 to BC; about the ▲ ABC describe (S. 5. 1. cor.) the circle ABC, and produce AD to meet the circumference in E; from DA, produced if necessary, cut off DF= DE; join B, E and C, E; also join B, F and C, F; and let BF and CF produced meet AC, and AB, in G and H respectively.

L

And since CD is common to the two CDF, CDE, and that (constr.) DF=DE, and the / CDF=▲ CDE, ... (E. 4. 1.) the ▲ FCD = 2 DCE or BCE; but (E. 21. 3.) the BAE= 2 BCE; ..the BAE or HAE FCD; and (E. 15. 1.) the AFH, of the A AHF, is also equal to the ▲ DFC, of the ▲ CDF; ... (S. 26. 1.) the ▲ AHF =2FDC, which (constr.) is a right ; .. the ZCHA is a right ; i. e. CFH is to AB; and, in the same manner, it may be shewn that BFG is to AC: Whence it is manifest, that the three perpendiculars cut each other in the common point F; for (E. 17. 1.) there cannot be drawn, from the same point, two different straight lines both of them perpendicular to the same straight line. 36. COR. The part of of any of the three perpendiculars, let fall from the three of a A, on the opposite sides, that is, between their common intersection and the circumference of the circle described about the A, is bisected by the side to which it is perpendicular.

PROP. XXVII.

37. PROBLEM. From either of the two given points in which two given circles intersect each other, to draw a chord cutting the one circumference, and meeting the other, such that the part of it, contained between the two circumferences, shall be equal to a given finite straight line.

Let the two given circles ABC, ABD, cut one

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another in the points A and B, and let L be a given finite straight line: From either of the two given points, as B, it is required to draw a straight line cutting either of the circumferences, as that of ABC, and meeting the other circumference, so meeting that the part of it contained between the circumferences, shall be equal to L.

Take any point C in the circumference of ABC, any point D in the circumference of ABD;

and

join A, B, and A, C, and A, D, and B, C and B, D; in AB, produced if necessary, take AFL; at the point A, in AF, make (E. 23. 1.) the ▲a FAGACB, and at the point F, make the < AFGADB, and let AG and FG meet in G. In the circle ABC place AH=AG; join B, H, and produce BH to meet the circumference of ADB in I: Then is HI=L.

For, join A, I: And, since (E. 22. 3.) the ▲ AHB+ 2 ACB=two right, and that (E. 13. 1.) the AHB + 2 AHI=two right , .. the <AHI ACB; but (constr.) the ACB= ZFAG; .. the AHI FAG; and (constr.) the AFGADB, which (E. 21. 3.) = 2 AIH;.. the AFG = ▲ AIB; and (constr.) the side AH, of the AHAI, is equal to the side AG, of the ▲ AGF; .. (E. 26. 1.) HI=AF; and (constr.) AFL; .. HI=L. Q. E. F.

PROP. XXVIII.

88. THEOREM. If two opposite angles of a quadrilateral figure be together equal to two right angles, a circle may be described about it.

Let any two opposite, as the ABC, ADC, of the quadrilateral figure ABCD, be together equal to two right: A circle may be described about the trapezium ABCD.

For, join A, C; and (S. 5. 1. cor.) about the A

B

DE

ABC describe a circle: Its circumference shall pass through the point D. If not, let it pass otherwise, so that, first, the point D is without the circle ABC, described about the A ABC; take any point E in the circumference of the circle and within the A ADC; and join A, E and C, E: Then, since ABCE is a quadrilateral figure inscribed in a circle the ABC + AEC=two right; and (hyp.) the ABC + ADC=two right; .. the AEC-2 ADC, which (E.21.1.) is absurd. Wherefore the point D is not without the circle ABC; and in the same manner it may be shewn that the point D is not within the circle ABC; .., the circumference of the circle ABC passes through the point D, and is, .., a circle described about the four-sided figure ABCD.

PROP. XXIX.

39. THEOREM. A circle cannot be described about a rhombus, nor about any other parallelogram which is not rectangular.

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