I found a simple, useful theorem with a very elegant proof :

Theorem: Consider a \( \Delta ABC \) with a point \( D \) on \( BC \) such that \( \angle BAD = \theta_{1} \) and \( \angle DAC = \theta_{2} \). If \( \overline{AB} = c \), \( \overline{AC} = b \) and \( \overline{AD} = p \), then we have the relation : \( \boxed{ p = \dfrac{bc\sin(\theta_{1}+\theta_{2})}{c\sin(\theta_{1})+b\sin(\theta_{2})} } \).

Proof: Area of \( \Delta ABD \) + Area of \( \Delta ADC \) = Area of \( \Delta ABC \) \( \Rightarrow \dfrac{cp\sin(\theta_{1})}{2} + \dfrac{pb\sin(\theta_{2})}{2} = \dfrac{cb\sin(\theta_{1}+\theta_{2})}{2} \) \( \Rightarrow p = \dfrac{bc\sin(\theta_{1}+\theta_{2})}{c\sin(\theta_{1})+b\sin(\theta_{2})} \).

With this theorem, one can easily find an elegant trig proof of Steiner Lehmus theorem.

Also check this out !

## Comments

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TopNewestBeautiful.. I'll trick my friends with this..ahhahahahaha Lol – Jun Arro Estrella · 1 year ago

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Given that the abstract says

It is not surprising that the proofs are more convoluted.

I believe that your proof is pretty standard for this result. – Calvin Lin Staff · 1 year ago

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– Karthik Venkata · 1 year ago

Can you link me any proof similar to mine ? By the way, I edited the link in the post to a much larger repository of Steiner Lehmus proofs.Log in to reply

@Karthik Venkata Is your solution not similar to the

decomposition into ABE and ADCgiven in the PDF? .... .WELL DONE THOUGH :) – Abhinav Raichur · 1 year agoLog in to reply

– Karthik Venkata · 1 year ago

Thanks for pointing it out, it is indeed using the same area decomposition technique. Yet, it doesn't really exploit the direct expression for angle bisector length.Log in to reply

It somehow reminds me of Cotangent Rule :P – Nihar Mahajan · 1 year ago

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– Karthik Venkata · 1 year ago

Hmmm not exactly. Cotagent rule relates \( \theta_{1} \) and \( \theta_{2} \) with \( \overline{BD} \) and \( \overline{CD} \).Log in to reply

– Nihar Mahajan · 1 year ago

Yes , thats why I said "somehow".Log in to reply