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(32.) If from any point within or without any rectilineal figure, perpendiculars be let fall on every the sum of the squares of the alternate segments made by them will be equal.

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Let ABCD be any quadrilateral figure (the demonstration being the same whatever be the number of sides). From any point I let perpendiculars IE, IF, IG, IH be drawn; AE2+ BFGC2+ DH2 = EB2 + FC2 + GD2 + AH2. From I draw lines to each of the angles; then AE2 + ER2 = (Al' =) AH2 + HIo, BF2 + FI2 = (BI2 = ) BE2 + EIo, CG + GI (CI2 = ) CF2 + FI2, DH2 + HI' = (DI2 = .) DG° + GI2,

whence,

=

AE+BF+CG2 + DH2 = EB2 + FC2 + GD*+HẨ3.

(33.) If from any point within a rectangular parallelogram lines be drawn to the angular points; the sums of the squares of those which are drawn to the opposite angles are equal.

Let ABCD be a rectangular parallelogram, and Fany point within it; join FA, FB, FC, FD; the squares of FA and FC are together equal to the squares of FB and FD.

D

B

Draw the diagonals AC, BD; and join FE. Because the triangles ADC, BDC are similar and equal, AC BD; and .. their halves, AE and DE, are equal.

Now (iv. 30.) FD2+FB2 = 2 DE2 + 2 EF3,

=2AE2+2EF2 = AF2+FC2.

(34.) The squares of the diagonals of a parallelogram are together equal to the squares of the four sides.

Let ABCD be a parallelogram, whose diagonals are AC, BD; the squares of AC, BD are together equal to the squares of AB, BC, CD, DA.

Since DB is bisected by AC,

D

2 AE2 + 2 ED2 = AD2 + AB2,

and for the same reason,

2 CE2 + 2 ED2 = CD2 + CB2,

.. 4AE+4ED' = AD2 + AB2+CB2+CD,
i. e. AC+BD2 = AD2 + AB2 + CB2+CDo.

B

(35.) If two sides of a trapezium be parallel to each other; the squares of its diagonals are together equal to the squares of its two sides which are not parallel and twice the rectangle contained by its parallel sides.

A

Let the sides AB, DC of the trapezium ABCD be parallel; draw the diagonals AC, BD; the squares of AC and BD, are together equal to the squares of AD and BC, and twice the rectangle AB, DC. Let fall the perpendiculars CE, DF.

B

Then (Eucl. ii. 12.), DB2=DA2+AB2+2AB × AF,

and AC CB + AB +2 AB × BE,

=

whence,

AC2+DB2 = AD2 + CB2 + 2AB2 +2AB × BE+2AB× AF.

Now (Eucl. ii. 1.),

AB× FE=AB × FA+AB× AB+AB × BE,

:. AC2 + DB2 = AD2 + CB2+2 AB × DC.

(36.) The squares of the diagonals of a trapezium are together double the squares of the two lines joining the bisections of the opposite sides.

Let ABCD be a trapezium, whose sides are bisected in E, F, G, H. Join EG, FH; and draw the diagonals AC, BD. The squares of AC, BD are together double of the squares of EG, FH.

H

E

B

Join EF, FG, GH, HE. Then (iv. 14.) EFGH is a parallelogram, and BD is double of EH;

.. BD2 = 4 EH2 = 2 EH2 + 2 FG3, and for the same reason AC2 = 2 EF2 + 2HG, ... AC2 + BD-2 EF2+2 FG+2 GH2+2 HE2, = 2 EG2+2 HF2. (iv. 34.)

(37.) The squares of the diagonals of a trapezium are together less than the squares of the four sides, by four times the square of the line joining the points of bisection of the diagonals.

Let ABCD be a trapezium whose diagonals AC, BD are bisected in E, F; join EF; the squares of AC,

BD are less than the squares of the four sides by four times the square of EF. Since BE bisects AC the base of the triangle ABC,

AB+BC2=2 AE2 + 2 EB2 ;

and for a similar reason,

AD2+ DC22 AE2 + 2 ED2;

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.. AB2+BC2+CD2 + DA2 = 4 AE2+2 EB2 + 2 ED2

= AC2+2 EB + 2 ED

= AC2 +4 BF2 +4 FE2

= AC + BD2 +4FE.

(38.) In any trapezium, if two opposite sides be bisected; the sum of the squares of the two other sides, together with the squares of the diagonals, is equal to the sum of the squares of the bisected sides together with four times the square of the line joining those points of

bisection.

Let AB, DC, two opposite sides of the trapezium ABCD, be bisected in E, and F; join EF; and draw the diagonals AC, BD. The squares of AD, BC, AC, BD are equal to the squares of AB, DC, and four times the square of EF.

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Join AF, BF. Since AF bisects DC the base of

the triangle ADC,

AD2 + AC2 = 2 DF2+2 FA;

and in the same manner,

BCBD = 2 DF2 + 2 FB2;

DF2

whence AD+BC2+AC2+BD2=4 DF2+2 FA2+2FB2 =DC2+2 FA2 + 2 FB2 = DC2+4AE +4 EF2 =DC2+AB2+4 EF2.

(39.) If squares be described on the sides of a rightangled triangle; each of the lines joining the acute angles and the opposite angle of the square, will cut off from the triangle an obtuse-angled triangle, which will be equal to that cut off from the square by a line drawn from the intersection with the side to that angle of the square which is opposite to it.

From the angles B, C of the right-angled triangle BAC, let lines BG, CD be drawn to the angles of the squares described upon the sides, and from the intersections H and I let HE, IF

D

E

I

B

be drawn to the opposite angles of the squares; the triangle BIC=AIF, and CHB = AHE.

Join AG, AD. Then (Eucl. i. 37.) the triangle AFI AIG; to each of which add ABI, .. the triangle BIF BAG = BCA (Eucl. i. 37.) From each of these equals take away the triangle BIA, and BIC= AIF. In the same manner it may be shewn that CHB =AHE.

(40.) If squares be described on the two sides of a right-angled triangle; the lines joining each of the acute angles of the triangle and the opposite angle of the square will meet the perpendicular drawn from the right angle upon the hypothenuse, in the same point.

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