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The words inscriptible, circumscriptible, escriptible mean capable of being inscribed in, circumscribed about, escribed to, a circle.

Exercise. 290. If any two chords cut within the circle, at right angles, the sum of the squares on their segments equals the square on the diameter.

PROPOSITION XVI.

202. Theorem. A circumference can be described to pass through the three vertices of any triangle. (Circumscribed circle.)

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Given

the points A, B, C, the vertices of ▲ ABC.

To prove that a circumference can be described to pass through

A, B, C.

Proof. 1. There is such a circumference.

§ 131, cor. 2

2. And the center of the ○ can be found. § 131, cor. 1

NOTE. The relation between prop. XVI and prop. XVII should be noticed. Similarly for props. XVIII and XIX, and for XX and XXI.

Exercises. 291. Prove from prop. XVI and prop. XI that the sum of the interior angles of any triangle equals a straight angle.

292. If the hypotenuse of a right-angled triangle is the diameter of a circle, the circumference passes through the vertex of the right angle. (COROLLARY. The median from the vertex of the right angle of a rightangled triangle equals half of the hypotenuse.)

293. A line-segment of constant length slides so as to have its extremities constantly resting on two lines perpendicular to each other. Find the locus of its mid-point.

294. If a circle is described on the line joining the orthocenter to any vertex, as a diameter, prove that the circumference passes through the feet of the perpendiculars from the other vertices to the opposite sides.

295. Prove that the perpendiculars from the vertices of a triangle to the opposite sides bisect the angles of the triangle formed by joining their feet; the so-called Pedal Triangle.

PROPOSITION XVII.

203. Theorem. A circle can be described tangent to the three lines of any triangle. (Inscribed and escribed circles.)

b

BR,

a

Given

the lines a, b, c, forming a ▲ ABC.

To prove that a circle can be described tangent to a, b, c.

Proof. 1. Let O be the in-center, 01, 02, 0, the ex-centers.

Let

Then

and

2.

OP, OQ, OR a, b, c.

AARO▲ AQO,

▲ BRO▲ BPO. I, prop. XIX, cor. 7

.. OQ = OR = OP.

3... P, Q, R are concyclic.

Why?

§ 108, def. O, cor. 3

4. And AB 1 OR, AB is a tangent.

Prop. IX, cor. 3

Similarly, a, b, c, are tangent to the other three .

COROLLARY. A circle can be described tangent to three lines not all parallel nor concurrent.

PROPOSITION XVIII.

204. Theorem. In an inscribed quadrilateral the sum or difference of two opposite angles equals the sum or difference of the other two opposite angles, according as the quadrilateral is convex or cross.

Given

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the inscribed convex quadrilateral ABCD.

To prove that in Fig. 1, ZA+ZC=ZB+ZD.

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$ 30 C-ZA

Proof for Fig. 2. If the quadrilateral is cross,

= ZD − ≤ B, since each equals zero. Why?

COROLLARIES. 1. A parallelogram inscribed in a circle has all of its angles equal, and is therefore a rectangle. (Why ?) 2. The opposite angles of an inscribed convex quadrilateral are supplemental.

Exercises. 296. In the figure of prop. XIII, if P is the mid-point of arc AB, prove that P is equidistant from AX and AB. Suppose the arc BCA is taken, instead of AB.

297. If a circle is described on one side of a triangle as a diameter, prove that the circumference passes through the feet of the perpendiculars drawn to the other two sides from the opposite vertices.

PROPOSITION XIX.

205. Theorem. In a circumscribed quadrilateral the sum or difference of two opposite sides equals the sum or difference of the other two opposite sides, according as the quadrilateral is convex or cross.

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Given

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the circumscribed convex quadrilateral abcd.

To prove that in Fig. 1, a + c = b+d.

Proof for Fig. 1, as lettered.

1. a1 = d2, a2 = b1, c1 = b2, c2 = d1. Prop. XIII, cor.

2. .. a1 + a2 + C1 + C2 = b1 + b2 + d1 + d2.

3. Or,

Ax. 2

Ax. 8

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Proof for Fig. 2. If the quadrilateral is cross, c — a = d — b.

cb2, and c2 = d1, .. c b2+ d1.

a + c = b+d.

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COROLLARY. A parallelogram circumscribed about a circle has all of its sides equal, and is therefore a rhombus. (Why?)

Exercises. 298. The bisector of an angle formed by a tangent and chord bisects the intercepted arc.

299. Given two pairs of parallel chords, AB|| A'B', and BC || B'C' ; prove that AC' || A'C.

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