In a ABC- the lengths of the bisectors of the angle A-B and C are x-y-z respectively- Show that 1 x cos A 2 - 1 y cos B 2 - 1 z cos C 2 - 1 a - 1 b - 1 c- Also show that a b-c - 1- x 2 bc

In ABC,x=2bcb+ccosA21xcosA2=12b+12c similarly 1ycosB2=12c+12a,1zcosC2=12a+12b 1xcosA2+1ycosB2+1zcosC2=12b+12c+12c+12a+12a+12b=1a+1b+1c √(1 x ...

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Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

In a ABC, t...

Question

In a $△ A B C$ , the lengths of the bisectors of the angle $A, B$ and $C$ are $x, y, z$ respectively. Show that $\frac{1}{x} cos \frac{A}{2} + \frac{1}{y} cos \frac{B}{2} + \frac{1}{z} cos \frac{C}{2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ . Also show that $\frac{a}{b + c} = \sqrt{1 - \frac{x^{2}}{b c}}$

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a + b + c = a b c

\frac{a (b^{2} c^{2} - 1)}{b c + 1} + \frac{b (c^{2} a^{2} - 1)}{c a + 1} + \frac{c (a^{2} b^{2} - 1)}{a b + 1} = 2 a b c

△ A B C, ∠ C = 90^{o}

p

C

A B

A B = c, B C = a, C A = b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively.

Show that :

$∠ A O B = \frac{1}{2} (∠ C + ∠ D)$

α, β, γ

\frac{1}{α} c o s \frac{A}{2} + \frac{1}{β} c o s \frac{B}{2} + + \frac{1}{γ} c o s \frac{C}{2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

a^{2}, b^{2}, c^{2}

\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}

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