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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestI 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?

Log in to reply

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

Log in to reply

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

Log in to reply