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# A geometry problem- I feel it is one

Hello! I have a doubt which came in my mind while solving a geometry question

Suppose we have 2 rays $$OA$$ and $$OB$$ with same end point $$O$$. Angle $$AOB$$ is any acute angle with value $$x$$.

another ray $$OC$$ is drawn such that it divides acute Angle $$AOB$$ into two parts, not necessarily equal.

Suppose there is a point $$D$$ on the ray $$OA$$.

My question is that, will a point $$E$$ always exist on ray $$OB$$ such that $$OC$$ bisect $$DE$$?

Note by Sarthak Singla
9 months, 1 week ago

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There will exist such a point $$E$$ only if $$\angle BOA + \angle BOC$$ is obtuse. · 9 months, 1 week ago

Can you give proof of your statement please · 9 months, 1 week ago

My proof is via complex numbers.

Without loss of generality assume that $$OB$$ represents the real axis.

Let $$D=\lambda e^{ix} ; \; \lambda \in \mathbb{R}^+$$.

Let $$D' (=\lambda' ; \; \lambda' \in \mathbb{R}^+)$$ be any point on $$OB$$.

Now, $$D'=E$$ if $\text{Arg} \left( \frac{D+D'}{2} \right) = y$ where $$y = \angle BOC$$

Simplifying the above condition, we get $\frac{\lambda'}{\lambda} = \sin x ( \tan y - \cot x)$

As the LHS is positive, so is the RHS.

Therefore,

$\tan y > \cot x$

Now, as $$0<y<x<90^\circ$$, we get (by using the cosine - sum and difference formulas)

$0>\cos (x+y)$

Due to the constraints on the angles $$x$$ and $$y$$, this again simplifies to $180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ$ · 9 months, 1 week ago

Thanks for the solution. Can it not be proved by some simpler method? I feel it would take me quite a few years to decipher this solution as I am unaware of complex numbers. · 9 months ago

I think E would also exist if angleBOA+angleBOC=90 · 9 months ago

It would be degenerate as $$E=O$$ in that case. · 9 months ago