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tenuse AB, the perpendicular distance of which from one of the sides, as AC, shall be equal to the segment of the hypotenuse between that point, and BC.

Bisect (E. 9. 1.) the ▲ ABC, by BD, and let BD meet AC in D; through D, draw DE (E. 31. 1.) parallel to CB: E is the point which was to be found.

For, since DE is parallel to CB, the ▲ BDE (E. 29. 1.); but (constr.) the

CBD=
CBD=

▲ DBE; ... ≤ DBE=✩ BDE; ... (E.6.1.) ED= EB; and since (hyp.) the C is a right, and that DE is parallel to CB, the ▲ CDE (E. 29. 1.) is a right; i. e. ED is I to AC.

PROP. XXIII.

33. PROBLEM. In the base of a given acute-angled triangle, to find a point, through which if a straight line be drawn perpendicular to one of the sides, the segment of the base, between that side and the point, shall be equal to the segment of the perpendicular, between the point and the other side produced.

Let ABC be the given acute-angled A: It is required, to find, in the base BC, a point through which if a perpendicular be drawn to AB, the segment of the base, between that point and the point

B, shall be equal to the segment of the perpendicular between that same point and AC produced. Draw (E. 11. 1.) from B, BY to AB; bisect

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(E. 9.1.) the

CBY by BD, meeting AC, pro

duced in D; through D, draw (E. 31. 1.) DE parallel to BY, and let DE cut BC in F: F is the point which was to be found.

For, since (constr.) the ABY is a right

▲, and that DE is parallel to BY, the E (E. 29. 1.) is, also, a right

; and the

YBD= YBD=4DBF; .,

BDF; .'. (E. 6. 1.) FB=FD.

▲ BDF; but (constr.) the DBF

the

PROP. XXIV.

34. PROBLEM. From a given isosceles triangle to cut off a trapezium, which shall have the same base as the triangle, and shall have its three remaining sides equal to each other.

Let ABC be the given isosceles ▲ It is re

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quired to cut off from it a trapezium, which, having BC for its base, shall have its three remaining sides equal to one another. Bisect (E. 9. 1.) the

ABC by BD, meeting AC in D; and through D draw (E. 31. 1.) DE parallel to CB: Then shall BE, ED, and DC, the three sides of the trapezium BEDC, be equal to one

another.

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For, since DE is parallel to BC, the ▲ AED= ABC (E. 29. 1.), and ADE=ACB; but (hyp. and E. 5. 1.) ABC= ACB; ..., 2 AED=2 ADE; .·. (E. 6. 1.) AE=AD; but (hyp.) AB=AC; from these equals take the equals AE and AD, there remains EB DC: Again, because DE is parallel to BC, the ▲ CBD= < BDE (E. 29. 1.); but (constr.) ▲ CBD= ▲ DBE; ... the DBE ▲ BDE;` .•. (E. 6. 1.) EB=ED; and EB has been proved to be equal to DC; ... EB, ED and DC are equal to one another.

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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 draw from A, a straight line which shall make with BC 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 < EAG=< AGC (E. 29. 1.); but (constr.) the LEAG=D; ... 4 AGC=✩ D.

PROP. XXVI.

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

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

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