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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 !

Note by Karthik Venkata
8 months, 1 week ago

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Beautiful.. I'll trick my friends with this..ahhahahahaha Lol Jun Arro Estrella · 8 months, 1 week ago

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Given that the abstract says

We offer a survey of some lesser known or new trigonometric proofs of the Steiner-Lehmus theorem.

It is not surprising that the proofs are more convoluted.


I believe that your proof is pretty standard for this result. Calvin Lin Staff · 8 months, 1 week ago

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@Calvin Lin 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. Karthik Venkata · 8 months, 1 week ago

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@Karthik Venkata @Karthik Venkata Is your solution not similar to the decomposition into ABE and ADC given in the PDF? .... .WELL DONE THOUGH :) Abhinav Raichur · 8 months, 1 week ago

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@Abhinav Raichur 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. Karthik Venkata · 8 months, 1 week ago

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It somehow reminds me of Cotangent Rule :P Nihar Mahajan · 8 months, 1 week ago

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@Nihar Mahajan Hmmm not exactly. Cotagent rule relates \( \theta_{1} \) and \( \theta_{2} \) with \( \overline{BD} \) and \( \overline{CD} \). Karthik Venkata · 8 months, 1 week ago

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@Karthik Venkata Yes , thats why I said "somehow". Nihar Mahajan · 8 months, 1 week ago

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