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Let BE, CF be squares described on the sides BA, AC containing the right angle. Join DC, BG; they

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intersect AL, which is perpendicular to BC, in the same point O.

Produce DE, GF, to meet in H. Join HA, HB, HC. Let BH, CH respectively meet DC, BG in I and K. Since EH-AF= AC, and EA=AB, and the angles HEA, BAC are right angles, the triangles HEA, BAC are equal, and the angle EHA=BCA=BAL, i. e. since EH and BA are parallel, HAL is a straight line, or LA produced passes through H, and HL is perpendicular to BC. Again, since AC=CG, AH = BC, and the angle HAC = BCG, .. the triangles HAC, BCG are equal; .. the angle CBK= CHL; but BCK =HCL; .. BKC=HLC, i, e. is a right angle, and BK is perpendicular to HC. In the same manner it may be shewn that CI is perpendicular to BH. Hence .. HL, CI, BK are perpendicular to the sides of the triangle HBC, and .. they intersect each other in the same point.

(41.) If squares be described on the three sides of a right-angled triangle, and the extremities of the ad

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jacent sides be joined; the triangles so formed are equal to the given triangle and to each other.

On the sides of the rightangled triangle ABC let squares be described, and join GH, FD, IE. The triangles AGH, BFD, ECI are equal to ABC, and to each other.

It is evident that AGH= ABC. Produce FB, and from D draw DS perpendicular to

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it. Since ABS and CBD are right angles, .. the angles ABC, SBD are equal; and BAC, BSD are also right angles, and BC=BD, .. DS=AC. And the triangles ABC, FBD being upon equal bases AB, FB are as their altitudes AC, DS (Eucl. vi. 1.); and .. are equal. In the same manner if IC be produced, and ER drawn perpendicular to it, it may be shewn that ER is equal to AB, and the triangle ECI to ABC. And since each of the triangles is equal to ABC, they are equal to one another.

(42.) If the sides of the square described upon the hypothenuse of a right-angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the legs; they will cut off triangles equiangular and equal to the given triangle.

Let DB, EC, the sides of the square described on BC the hypothenuse of the right-angled triangle ABC, be produced to meet the sides of the squares described

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In like manner it may be proved that the triangles ABC, LCI are equiangular and equal.

squares de

(43.) If from the angular points of the scribed upon the sides of a right-angled triangle perpendiculars be let fall upon the hypothenuse produced; they will cut off equal segments; and the perpendiculars will together be equal to the hypothenuse.

Let FM, IN be drawn from

the angles F, I of the squares described upon BA, AC, perpendicular to BC the hypothe

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nuse produced; MB will be M B

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equal to NC; and FM, IN together equal to BC.

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From A draw 40 perpendicular to BC. Since FBA is a right angle, the angles FBM and ABO together are equal to FBM and BFM, :. ABO is equal to BFM; and the angles at M and O are right angles, and AB= BF, .. BM=AO, and FM=BO. In the same manner may be shewn that CN=AO, and IN=CO; .. MB

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=NC, and FM and IN together are equal to BO and CO together, i. e. to BC.

COR. The triangles FBM, ICN are together equal to ABC.

(44.) If on the two sides of a right-angled triangle squares be described, the lines joining the acute angles of the triangle and the opposite angles of the squares will cut off equal segments from the sides; and each of these equal segments will be a mean proportional between the remaining segments.

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On AB, AC the sides of the right-angled triangle BAC, let squares be described, and BI, CF joined; the segments AP, AQ are equal, and each of them is a mean proportional between BP and CQ. Since AQ is parallel to HI, and AP to FG, BH : HI :: BA : AQ,

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B

and (CA) HI: CG AP: (FG=) AB,
.. BH CG:: AP AQ;

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and BH being equal to CG, AP=AQ.

Again, the triangles BPF, ACP being similar, as also

ABQ, ICQ,

BP: (BF) AB :: AP

AC,

and BA : AQ :: (IC=) AC : CQ,

.. ex æquo BP : (AQ=) AP :: AP: CQ.

(45.) If squares be described on the hypothenuse and sides of a right-angled triangle, and the extremities of

the sides of the former and the adjacent sides of the others be joined; the sum of the squares of the lines joining them will be equal to five times the square of the hypothenuse.

Let squares be described on the three sides of the rightangled triangle ABC; join DF; EI; the squares of DF and EI together are equal to five times the square of BC.

Draw FK, IL perpendicular to DB, EC produced, and AM to BC. The angle FBK is

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equal to ABC, and the angle at K to the right angle AMB, and FB=BA, .. BK=BM.

CL=CM.

In the same way,

Now (Eucl. ii. 12.) FD2 = DB2 + BF2 + 2 DB × BK

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... FD2 + EI2 = 2 BC2 + BA2 + AC2 + 2BC × BM+

[2 BCX CM

=2 BC+BC+2 BC=5 BC.

(46.) If a line be drawn parallel to the base of a triangle, and terminated in the sides; to draw a line cutting it, and terminated also by the sides, so that the rectangle contained by their segments may be equal.

Let ED be parallel to CB the base of the triangle ABC; from D draw DF, making with AC (produced if

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