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Cor. 1. This is another proof that the nine-points circle touches the incircle.

CoR. 2. If we write I, for I, r, for r, and + for in the above we have the similar proposition for the escribed circle.

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Join AK. Through K draw By antiparallel to BC. Join SA cutting By at right-angles in .. Then the triangle ABy is similar to ABC; so that

Αλ

AK By _2p

AL AD BC
Thus SKP = SA? + KA2 – 2SA.LA

4p2

AD2 2R. a

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a

2p

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Cor. 1. Square on diameter of Brocard Circle = Ro – 3p?.

Cor. 2. Since K and S'K subtend angles equal to w in the Brocard Circle,

:: NK = N'K = SK sin w= R sin w 1 (1 – 3 tan’ w), and N2S = N'S = SK cosw=R J(cos w – 3 sino w),

and 12.02 SK sin 2w = R sin 2w J(1 – 3 tan’ w). This last result was reached in Art. 215.

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222. Area of a trapezium (i.e. a quadrilateral with one pair of opposite sides parallel).

Let ABCD be a quadrilateral in which AB, CD are parallel. Then ABCD=

= A ABD + ABCD = * (AB + DC) (perpendicular distance between AB and CD), = } (sum of parallel sides) altitude.

A

B
This
may

be written
(AB+CD) AD sin ADC or (AB+CD) BC sin BCD.

223. Area of a quadrilateral inscribable in a circle.

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.. adding and subtracting

2 (ad + bc) (1 + cos A) = (a + d)2 (6 -c),
2 (ad + bc) (1 - cos A)=(b + c)2 (a - d),

.. multiplying
4 (ad + bc)? sino A

= (a +d+b-c)(a + d b+c) (b+c+a-d) (b+ca + d), .. dividing by 16 and taking the square root

Quadrilateral ABCD= {(8 a) (8 6) (8c)(8 d)}......(2).

224. The area of any quadrilateral (in terms of its sides and diagonals).

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Let Q= area of ABCD.
AB= a, BC = b, CD=c, DA = d, AC = , BD=f.

Let AC, BD cut in 0. Then
Q=A0B + BOC + COD + DOA
=} (40.0B sin AOB + BO.OC sin BOC

+CO. OD sin COD + DO.OA sin DOA) = 1 AC. BD sin AOB = } ef sin AOB

(1). Now a” = A02 + BO2 2 A0. BO cos AOB,

62 = BO2 + CO2 2BO.CO cos BOC,
C = CO2 + DOP 200. DO cos COD,
ď? = D02 + A02 2DO. AO

cos DOA. :. - a+ d - c = 2 (10.0B + BO.OC

+ CO.OD + DO.0A) cos AOB 2ef cos AOB. :. 2ef (1 + cos AOB) = 2ef + b2 + d - a - c?,

2ef (1 – cos AOB)=2ef + a2 + C2 62 d, :. 16Q2 = (2ef + b2 + d2 ao co) (2ef + ao + c2 62 - d)... (2). Cor. 1. Since in a circle efrac + bd, (Euc. VI. D) therefore his equation reduces to the formula (2) of the preceding article. Cor. 2. If a circle is inscribable in the Quadrilateral,

a+c=b+d,
:: Q=1 /{ef + bd ac) (ef bd + ac)}.

a

225.

The area of any quadrilateral: (in terms of its sides and the sum of its opposite angles.)

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Let A + C= 20, so that B+D=2 (). Then

Q=A ABD+ ABCD = fad sin A + bc sin C.......(1).

Now

BD2 = a? + d2 2ad cos A = 12 + C2 - 2bc cos C.

.. a- - 62 +d- = 2ad cos A – 2bc cos C.

=

:. (a? 62 + d2 – 2)2 = 4a ́do coso A Babcd cos A cos C + 46% cosa C, and

16Q2 = 4aRd2 sino A + Babcd sin A sin C + 46%c- sino C. :. 16Q2 + (a? 82 + D2 - 2)2 = 4aBabcd cos (A + C) + 4boco. .. 16Q* = 4 (ad + bc)2 – (a? 62 + ď? - c^)2 – 16abcd cos? O = (2ad + 2bc + ab2+do -c) (2ad + 2bc - a? + 62 - + c)

- 16abcd cosa 0. :: Q2 = (8 a) (8 6) (8-c) (8 d) abcd cosa 0...... (2).

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CoR. 1. If Q is inscribable in a circle, 20= , 0 = 1 and cos 0 = 0. Hence Q= {{(s- a) (8 b) (8 c) (8 d)}.

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CoR. 2. If a circle is inscribable in Q,

8 - a=C, 8–6 = d, 8-c=Q, 8 - d=b.

..Q=/(abcd). sin 0.

=

=

Cor. 3. A quadrilateral of given sides has its maximum area when it is inscribable in a circle.

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