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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.

Note by Nihar Mahajan
2 years, 5 months ago

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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?

- 2 years, 5 months ago

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

- 2 years, 5 months ago