PROP. 36, THEOR.

If from a point without a circle two right lines be drawn to it, one of which is a tangent to the circle and the other cuts it, the rectangle under the whole secant and the external segment is equal to the square of the tangent.

1. Let the secant pass through the centre, draw a right line from the centre to the point of contact. The 02 2 of the tangent is to the difference between the O'rs of the segment (between the centre and given point) and the radius, (prop. 47, b. 1,) but the rectangle under thewhole secant and external segment is to the same difference, (prop. 6, b, 2,).. &c.

2. If the secant does not pass through the centre, connect the centre with the given point, with the extremities of the given lines and with the point where the secant cuts the circumference.

Then the rect. under the whole secant and external segment is to the dif. between the rs of the line from the centre to the given point and radius, (cor. prop. 6, b. 2,) but the of the tangent is also this difference, ... &c.

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Cor. 1. Hence if from any point without a circle, two right lines be drawn cutting the circle, the rectangles under them and their external segments are, for each of the rectangles is to the ' of the tangent.

Cor. 2. If from the same point two lines be drawn to a circle, which are tangents to it, they are; for their squares are equal to the same rectangles.

PROP. 37, THEOR.

If from a point without a circle two, right lines be drawn, one cutting the circle and the other meeting it, and if the rectangle under the secant and external segment be equal to the square of the line which meets the circle, the line which meets it is a tangent.

Draw from the point without, a tangent to the circle, and connect the centre with the given point, the point of contact and the extremity of the meeting line.

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The of the tangent is to the rect. under the secant and external segment, and also the 2 of the meeting line is to the same rect... these 2rs are =, and.. the lines themselves; then in the triangles formed by the radii, the line from the centre to the point of contact and those lines, the three sides of the one are respectively to those of the other, .. the angle contained by the tangent and radius is to that contained by the meeting line and radius, (prop. 8, b. 1,) .. the meeting line is a tangent, (prop. 16, b. 1.)

OBSERVATIONS.

1. To cut a given right line [AB] so that the rectangle under the parts shall be equal to any square less than that of half the given line.

Fig. 6.

ANALYSIS.

Suppose it done, and that the rectangle under AC,CB is to any given less than the of AB: bisect AB and describe on it a semicircle; from the point C raise a perpendicular CD and join DA and DB. Then DAB is a right angled triangle, and the rect. ACB is= to the of CD, (note prop. b. 2,) .. CD is to the given 2, and .. CD is to its side.

2

Therefore describe on the given line a semicircle; from either extremity B draw a perpendicular BE to a side of the given; draw ED parallel to BA, and from the point D in which it meets the circumference, let fall a perpendicular DC; it evidently cuts the given line in the required point.

2. To produce a given right line [AC] so that the rectangle under the whole produced line and added part may be equal to any given square.

Suppose it done, and that CD is the produced part; bisect the given line in B, then since AC is bisected in B and produced to D, the rect. under AD,DC with BC2 is to the 2 of BD, ... BC2 is the difference between the given and BD; .. from either extremity C of the given line raise a perpendicular, and inflect from the point of bisection B a right line BE to BD, then EC is evidently to the side of the given 2.

Therefore, draw from either extremity C of the given line a perpendicular CE to a side of the given 2, bisect AC in B, join BE and produce AC 'till BD is = to BE; the rect. under AD,DC is to the given

2

2

For the rectangle ADC with the of BC is to the of BD, (prop. 6. b. 1, of Elr.) but the of BC is the difference between the 'rs of BE (or BD) and CE, .. the rectangle ADC is to the of CE and.. to the given ☐2.

3.

If in a triangle [HGI] the rectangle under the seg ments [GK,KH} of one side made by a perpendicular [K] from the opposite angle, be equal to the square of that perpendicular, the angle [GIH] from which it is drawn, is a right angle.

Suppose that the angle GIH is a right angle; then the of GH is to the sum of the rs of GC,CH, if .. this can be proved, the proposition is true. Because the rect GKH is to the of KI, and KI? is the difference between CH and KH', . the rect. GHK is to the 2 of HI; also the rect. HGK is to the 2 of GI, but those rectangles are together to the 201 GH, (prop. 2, b. 2 of Elr.) .. GIH is a right angle, •. &c.

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4. To produce a given right line [GK] so that the rectangle under it and the produced part may be equal to any given square.

Suppose it done, and that the rectangle GKH is to the given quantity, KH is the produced part; draw KI at right angles to GH, join GI and HI. Then because the rectangle GKH is to KI2, the angle GIH is right, (prop. 4, b. 2, dedu.)

Therefore draw from either extremity K of the given line, a perpendicular KI to a side of the given, join GI, draw through I, IH at right angles to IG and produce GK to meet it, KH is evidently the required produced part.

Given a seelene triangle [ABC] to produce one side [CB] so that the rectangle under it and the produced, part may be equal to the difference of the squres of the other two sides.

Suppose it done, and that BD is the produced part; join AD, it is to the side AC; for if not, draw any

T

other right line AG [= to it] to meet the produced part; then, because AC and AG are, the rectangle under CB,BG is to the dif. between the 'rs of AB and AC, (cor. prop. 1, b. 2,) but the rect. CBD is also to it, which is absurd,.. AG is not to AC, nor is there any other line but AD to it.

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Therefore, from A as a centre, with the radius AC describe a circle and produce CB to meet its circumference; the rect. CBD is to the difference between the 'rs of BA, AC, (cor, prop. 6, b. 1.)

6. The square of the greater of any two lines, is equal to four rectangles under the less and difference, with a square of the difference between the less and difference.

2

For the of the greater is to a 2 of the less, a2 of the difference and two rectangles under the less and difference, but the rs of the less and difference are together to two rectangle under the less and dif. with a of the difference between the less and difference.

7. The square of one side of a triangle cannot be equal to the sum of the squares of the other two and two rectangles under them; for if so, this side would be equal to the other two.

8. To cut a given right line [AB] so that the sum of the squares of the unequal parts may be equal to any quantity greater than two squares of half, and less than the square of the given line.

Suppose it done, and that the rs of AD,DB are together to the given quantity; bisect the given line at C; then two rs of AC with two 'rs of CD are

to the sum of the rs of AD and DB, (prop. 9, b. 2,).. the of CD is to half the difference between the given quantity and two'rs of half the given line.