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 \(DE\)?

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## Comments

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TopNewestThere will exist such a point \(E\) only if \(\angle BOA + \angle BOC\) is obtuse.

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Can you give proof of your statement please

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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 \[180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ\]

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@Sarthak Singla

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I think E would also exist if angleBOA+angleBOC=90

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It would be degenerate as \(E=O\) in that case.

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