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

Cor. 2. If we write 1, 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 l. Then the triangle ABy is similar to ABC; so that

Αλ ΑΚ βγ

AL AD BC
Thus SK? = SAP + KA2 – 2SA LA

4p

. AD - 2R. 24.LA a?

(26°+ 2c* – a* – 2a* – 26° – 2c*) [for 4AD2 = 262 + 2c – a,

a. LA = 2S,

4R.S = (a* + b2 +co) p],
.. SK = Ro - 3p = Ro (1-3 tanow).
J. T.

11

= R+

Cor. 1. Square on diameter of Brocard Circle = R2 – 3p.

COR. 2. Since 2K and NK subtend angles equal to w in the Brocard Circle,

.. NK = 12'K = SK sin w= -R sin w „(1 – 3 tanow), and 12S = D'S= SK cos w =R J(coso w - 3 sin’w), and NN' = SK sin 2w = R sin 2w 1(1 – 3 tan’ w).

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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

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

A

B
This

may be written
1 (AB+CD) AD sin ADC or 1 (AB+CD) BC sin BCD.

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i.e.

2 (ad + bc) cos A = a2 + d62 – c, and

2 (ad + bc) 1 = 2ad + 2bc. .. adding and subtracting

2 (ad + bc) (1 + cos A)=(a + d)2 – (-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 + c = a + d), .. dividing by 16 and taking the square root

+

Quadrilateral ABCD= {{(s 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 = 6, CD=C, DA = d, AC = e, BD=f.

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

+ CO.OD sin COD + DO.0A sin DOA) = AC.BD sin AOB = } ef sin AOB

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

62 = BO2 + CO2 2BO.CO cos BOC,
c = CO2 + DOC 2C0. DO cos COD,

do = D0% + 402D0.40 cos 00 4. .: 62 a2 + d2 -0= 2 (10.0B + BO.OC

+ CO. OD + DO.0A) cos AOB = 2ef cos AOB. .. 2ef (1 + cos A OB) = 2ef + b2 + d-ao-c",

2ef (1 - cos AOB)=2ef + a2 +62 - 62 - d, :: 16Q2 = (2ef + b2 + d - a - c)(2ef + a +c9-62 - 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)}.

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

D

A

Let A + C = 20, so that B+ D= 2 (** 0). Then

Q=A ABD +BCD = {ad sin A + 1 bc sin C.......(1).

Now

BDP = a + d2 2ad cos A = 3a + cé — 2bc cos C.

.. a- - 72 + d2 - C = 2 ad cos A – 2bc cos C.

+

.. (a? 62 + d2 – c2)2 = 4a ́da cosA – Babcd cos A cos C + 462ccosC, and

16Q2 = 4a ́d? sino A +8abcd sin A sin C +46c sin’ C. :: 16Q2 + (a? 62 +-co? = 4aod - 8abcd cos (A + C) + 4boco. .. 16Q2 = 4 (ad + bc)– (62 +d? — c)2 – 16abcd cosa 0 = (2ad + 2bc + a? b2 + d? c) (2ad + 2bc a2 + 62 d+co)

-16abcd cosa 0.

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

CoR. 1. If Q is inscribable in a circle, 20 = 1, 0 = 1 and cos 6 = 0. Hence Q= {{(s a) (8 6) (8 — c) (8 – d)}.

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

8 - arc, 8-b=d, 8-c= = A, 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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