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PROP. XXV.

35. PROBLEM. To draw to a given straight line, from a given point without it, another straight line which shall make with it an angle equal to a given rectilineal angle.

Let BC be a given straight line, A a given point

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without it, and D a given rectilineal : It is required to dray from A, a straight line which shall make with an equal to the

D.

Through A draw (E. 31. 1.) EAF parallel to BC; at the point A in EAF, make (E. 23. 1.) the LEAG=D: AG, is the line which was to be

drawn.

For, since (constr.) EF is parallel to BC, the LEAG=4 AGC (E. 29. 1.); but (constr.) the LEAG=D; .'. 4 AGC=4D.

PROP. XXVI.

36. THEOREM. If all the angles but one of any rectilineal figure, be together, equal to all the

1

angles but one, of another rectilineal figure having the same number of sides, the remaining angle of the one figure, shall be equal to the remaining angle of the other: And, conversely, if an angle in the one figure be equal to an angle in the other, the remaining angles of the one shall be equal, together, to the remaining angles of the other.

For, since the two figures have the same number of sides, all the interior

gether, equal (E. 32. 1. cor. 1.)

of the one are, to

to all the interior

of the other: If... from these equals be taken first, the aggregates which, by the hypothesis, are equal; and secondly, the single angles, which are supposed to be equal, it is manifest that the remaining angle, or angles, of the one figure, must be equal to the remaining angle, or angles of the other.

PROP. XXVII.

37. THEOREM. The angle at the base of an isosceles triangle is equal to, or is less, or greater, than the half of the vertical angle, accordingly as the triangle is a right-angled, an obtuse-angled, or an acute-angled triangle.

For, (E. 5. 1, and E. 32. 1.) the double of the

at the base

the vertical = two right

;

if .. the vertical be a right 4, and if it be taken from both, there remains the double of the a right ; ... the at the base

▲ at the base

half of a right ▲.

But, if the vertical be obtuse, when it is taken away from the same equals as before, there will remain the double of the 4 at the base equal to a less than a right ; ... the at the base is, in this case, less than the half of a right .

And, in like manner it may be shewn, that, when the vertical of the isosceles ▲ is acute, the at the base is greater than the half of a right angle.

PROP. XXVIII.

38. THEOREM. If either of the equal sides of an isosceles triangle be produced, towards the vertex, the straight line, which bisects the exterior angle, shall be parallel to the base.

at

For, (E. 5. 1. and E. 32. 1.) the exterior the vertex of an isosceles A is the double of either of the at the base; .. the half of that interior is equal to either of the at the base; ... the straight line bisecting the vertical is (E. 28. 1.) parallel to the base.

PROP. XXIX.

39. THEOREM. The distance of the vertex of a triangle from the bisection of its base, is equal to, greater than, or less than the half of the base, accordingly as the vertical angle is a right, an acute, or an obtuse angle.

First, let ABC be a right-angled A, right

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angled at A, and let AD join A and the bisection, D, of the base: AD=DB, or DC.

For, if not, AD is either greater or less than BD: Produce BA to X; and first, if it be possible, let AD > DB; ..., also, AD>DC; ... (E. 18. 1.) <B> <BAD, and <C> < CAD; .. 2 B+ C> < BAD+ ≤ CAD ; i. e. ▲ B+ ≤ C> < BAC; but (hyp. and E. def. 10. 1.) <BAC=CAX; .. < B+ <C> < CAX; which (E. 32. 1.) is

absurd.

And, in like manner, if DA be supposed to be less than BD, it may be shewn that 4 B+ < C <

¿CAX; which is absurd. Therefore, DA-DB, or DC.

Next, let the vertical / CEB, of the A EBC, be acute, and let ED join E and the bisection, D, of BC, ED> BD, or DC.

and, since (constr.)

From either of the B or C, as C, if the A EBC be acute-angled, draw (E. 12. 1.) CA L to the opposite side EB; and join A, D: Then (S. 7. 1.) CA falls within EB; the CAE is a right, the than a right ; .. (E. 17. 1.) the than a right, and.. less than the DAE; .. (E. 19. 1.) DE> DA; but, by the former case, DA=DB; ... DE> DB, or DC.

DAE is greater

AED is less

L

Lastly, if FBC be an obtuse-angled A, obtuseangled at F, join F,D; draw, as before, CA 1 BF; and join A, D: Then (S. 7. 1.) CA falls without BF, and the AFD (E. 16. 1.) > the FBD; but since (1st case) DA= DB, the DBF L DAF (E, 5. 1.); .'. ▲ AFD > ≤ DAF; .. (E. 19. 1.) DA > DF; but DA=DB; .. DF< DB, or DC. Or, the two last cases may be proved, ex absurdo, in the same manner as the first is proved.

40. COR. 1. If any number of triangles have a right angle for their common vertical angle, and have equal hypotenuses, the locus of the bisections of the several hypotenuses is a quadrantal arch of a circle, having the common vertex for its centre, and the half of any hypotenuse for its radius.

For, the bisections of the hypotenuses will, each of them, (S. 29. 1.) be at a distance from the

D

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