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# Check this out!

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 · 8 months, 1 week ago

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. Staff · 8 months, 1 week 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. · 8 months, 1 week ago

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

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

Hmmm not exactly. Cotagent rule relates $$\theta_{1}$$ and $$\theta_{2}$$ with $$\overline{BD}$$ and $$\overline{CD}$$. · 8 months, 1 week ago