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and (E. 15. 1.) the BDE 4 ADC ; .·. (E. 4. 1.) BE AC. But (E. 20. 1.) AB + BE > AE; but AC has been proved to be equal to BE, and AE is (constr.) the double of AD; ... AB+ AC> 2 AD...

20. THEOREM.

PROP. XIII.

The two sides of a triangle are, together, greater than the double of the straight line drawn from the vertex to the base, bisecting the vertical angle.

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Let ABC be any given A, and let AD be drawn from the vertex A, to the base BC, bisecting the vertical BAC: Then, AB + AC > 2 AD.

If the given ▲ be isosceles, the straight line which bisects the vertical is (E. 4. 1.) 1 to the base; and since (E. 17. 1. and E. 19. 1.) each of the equal sides is greater than the perpendicular, the proposition, is, in this case, manifestly true.

But, let ABC be a scalene A, and let the side AB be less than AC: Then, of the segments into which AD, bisecting the 2 BAC, divides the

base BC, BD, which is adjacent to the less side AB, is the less.

For, from AC, the greater, cut off (E. 3. 1.) AE= AB, the less, and join D, E; and because BA, AD are equal to EA, AD, and (hyp.) the < BAD= < EAD, ... (E. 4. 1.) BD=DE, and 2 BDA = 2 EDA; but (E. 16.1.) ≤ DEC > L ADE; ... ¿ DEC > ▲ ADB; and (E. 16. 1.) 2 ADB > < ACD; much more then is DEC> < ECD; ... (E. 19. 1.) DC > DE; but it has been shewn that DE=DB; ... DC> DB. From DC, the greater cut off (E. 3. 1.) DF=DB; and join A, F: Then (E. 16. 1.) the AFC > < ABC; and because (hyp.) AC > AB, (hyp.) AC > AB, .'.(E. 18. I.) 4 ABC > < ACB; much more then is AFC> 4 ACF;... (E. 19. 1.) AC > AF: But (S. 12. 1. and constr.) AB+ AF>2AD; much more then is AB+AC>2 AD.

21. COR. From the demonstration it is manifest, that of the segments into which the straight line bisecting the vertical of a scalene A, divides the base, that which is adjacent to the less side, is the less.

PROP. XIV.

22. THEOREM. If a trapezium and a triangle stand upon the same base, and on the same side of it, and the one figure fall within the other, that which has the greater surface shall have the greater perimeter.

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Let the trapezium EBCF fall within the. Á DBC; let, also, the ▲ DBC fall within the trapezium ABCD; and let all the figures stand on the same base BC: The perimeter of the A DBC is the perimeter of EBCF, and < the perimeter of ABCD.

First, let E and F be in the sides DB and DC of the A DBC, and let the vertex D of the ▲ DBC coincide with the A or the D of the trapezium ABCD.

Then, since (E. 20. 1.) DE+DF > EF, add to both, EB,BC, and CF; ... DE+EB+DF+FC +BC>EF+FC+CB+BE; i. e. the perimeter of the A DBC > the perimeter of EBCF.

Again, since (E. 20, 1.) BA+AD > BD, add to both DC and CB; ... BA+AD+DC+CB> BD+DC+CB; i. e. the perimeter of the trapezium ABCD > the perimeter of the ▲ DBC.

And, if E or F fall within the ▲ DBC, and the vertex of the A do not coincide with either of the A or D, of the trapezium, it may, in the same manner, be proved, that the proposition is true, a fortiori.

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23. PROBLEM. One of the angles at the base of a triangle, the base itself, and the aggregate of the two remaining sides, being given, to construct the triangle.

Let K be the given angle, AB the given base

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of the triangle, and H the aggregate of the two remaining sides: It is required to construct the triangle.

At the point A, in AB, make (E. 23. 1.) the BACK, and make (E. 3. 1.) AC=H; join C, B; and at the point B, in CB, make (E. 23. 1.) the CBD ACB: Then is DAB the triangle which was to be constructed.

For, since (constr.) DCB= 2 DBC, ..

C

E. 6. 1.) BD=DC; add to both DA; BD+DA=CD+DA; i. e. BD+DA=CA; and (constr.) CA=H; ... BD+DA=H; and the ZA was made equal to the given K: It is manifest, therefore, that DAB is the triangle which was to be constructed.

PROP. XVI.

24. THEOREM. If two right-angled triangles have the three angles of the one equal to the three angles of the other, each to each, and if a side of the one be equal to the perpendicular let fall from the right angle upon the hypotenuse of the other, then shall a side of this latter triangle be equal to the hypotenuse of the former.

Let ACB and EDF be two right angled A,

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right angled at C and D, having, also the < DEF 4 ABC, the EFD CAB, and the side AC, of the A ABC, equal to the perpendicular DG, drawn D to the hypotenuse EF of the A DEF: The side DE, of the A

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