Help/hint please

If $ABCD$ is a cyclic quadrilateral , prove that $\tan^2\dfrac{B}{2} = \dfrac{(s-a)(s-b)}{(s-c)(s-d)}$

$s$ is semi-perimeter of quadrilateral.

#Geometry

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Comments

I don't have a straight forward solution for this at the moment, I'll try something different later.

Essentially, it suffices to express the tangent in terms of the side lengths. Working on a triangle $ABC$ first, we can easily obtain that $\tan^2\frac {A}{2}=\frac {(s-b)(s-c)}{s(s-a)}$ where variables pertain to $ABC$ . Transferring this to $ABCD$ , we have $\tan^2\frac {A}{2}=\frac {BD^2-(b-c)^2}{(c+d)^2-BD^2}$ . I then used symmetry given by $\tan^2\frac {C}{2}$ to find that $BD^2=\frac {(bd+ac)(ab+cd)}{(cb+ad)}$ (This formula actually gives rise to another problem of ptolemy's theorem). I guess it is just manipulations from now on to obtain the desired RHS.

btw what is the conventional way to decide which side is $a,b,c,d$ for a quadrilateral?

Self note: the three quantities $(bd+ac),(ab+cd),(cb+ad)$ seem to play a big role in lengths associated with quadrilaterals. Maybe I could formulate a problem out of them?

Xuming Liang - 5 years, 10 months ago

Thanks a lot dude! I managed to complete it :)

Nihar Mahajan - 5 years, 10 months ago

@Calvin Lin @Pi Han Goh @Chew-Seong Cheong @Xuming Liang

Nihar Mahajan - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$