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PROPOSITION XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled A, having the obtuse LACB, and from A let AD be drawn / BC produced.

Then sq. on AB shall be greater than sqs. on AC, CB, by twice the rect. BC, CD.

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Dem. :: BD is divided into two parts at C, .. sq. on BD = sqs. on BC, CD, and twice rect. BC, CD;

(ii. 4) to each of these =s add sq. on DA; .. sqs. on BD, DA = sqs. on BC, CD, DA and twice

rect. BC, CD; but sq. on BA= sqs. on BD, DA, and sq. on AC on CD, DA;

(i. 47) :: sq. on BA = sqs. on BC, AC, and twice rect. BC, CD;

that is, sq. on BA is greater than sqs. on BC, AC, by twice rect. BC, CD. Therefore, in obtuse-angled triangles, &c. Q. E. D.

= sqs.

1. The perpendiculars from two angles A, B of an equilateral triangle on the opposite sides intersect in O: show that the square on AO is a third of the square on AB.

2. In the triangle ABC each of the angles at B, C is double the angle at A: show that the square on AB is equal to the square on BC together with the rectangle AB, BC.

PROPOSITION XIII. THEOREM.

In every triangle, the square on the side subtending an alute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.

Let ABC be any A, and LB one of its acute s, and upon BC, produced if necessary, let fall the I AD from the opp. L; then sq. on AC opp. _B shall be less than sqs. on CB, BA, by twice rect. CB, BD.

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sq. on CD.

Dem. For in fig. 1, BC is divided into two parts in D,

and in fig. 2, BD is divided into two parts in C, i, in both, sqs. on BC, BD = twice rect. BC, BD and

(ii. 7) Add to each of these =s sq. on DA, then sqs. on BC, BD, DA = twice rect. BC, BD and sqs.

on CD, DA; but sq. on AB = sqs. on BD, DA ; sq. on AC : = sqs. on CD, DA;

(i. 47) .. sqs. on BC, AB = twice rect. BC, BD and sq. on AC; ., sq. on AC is less than sqs. on AB, BC by twice rect.

BC, BD.

The case when AC is I BC needs no proof. (i. 47)

Therefore, in every triangle, &c. Q. E. D. 1. DE is parallel to the base BC of an isosceles triangle ABC. Show that the sq. on BE is equal to the rectangle contained by BC, DE, together with the sq. on CE.

2. If D be a point in the side BC of an equilateral triangle ABC prove that the squares on BD, CD are together less than the square on AD by the rectangle contained by BD, CD.

PROPOSITION A. THEOREM.

In every triangle, the sum of the squares on two of the sides, is equal to twice the square on half the base, together with twice the square on the line drawn from the vertex to the middle point of the base.

Let ABC be A and let D be the middle point of the base AB. Then shall the sqs. on AC, BC = twice the sq. on AD and twice the sq. on CD.

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Constr. Draw CE I AB.
Dem. Of the two LS ADC, BDC one must be acute

and the other obtuse, or each is a rt. L.
The case when each is a rt. Z needs no proof. (i. 47)
Let LADC be an obtuse L.

Then sq. on AC = sqs. on AD, DC with twice rect. AD, DE;.

(ii. 12) sq. on BC with twice rect. BD, DE = sqs. on BD, DC;

(ii. 13) .: sqs. on AC, BC, with twice rect. BD, DE = sqs. on AD, DB, and twice sq. on DC and twice rect. AD, DE;

(ax. 2) but rect. BD, DE=rect. AD, DE, and sqs. on AD, DB = twice sq. on AD, .: AD= DB;

(constr.) .. sqs. on AC, BC = twice sq. on AD and twice sq. on DC.

Therefore, in every triangle, &c. Q. E. D.

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