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 \(DE\)?

## Comments

Sort by:

TopNewestThere will exist such a point \(E\) only if \(\angle BOA + \angle BOC\) is obtuse. – Deeparaj Bhat · 3 months ago

Log in to reply

– Sarthak Singla · 3 months ago

Can you give proof of your statement pleaseLog in to reply

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 \[180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ\] – Deeparaj Bhat · 3 months ago

Log in to reply

@Sarthak Singla – Deeparaj Bhat · 3 months ago

Was this helpful?Log in to reply

– Sarthak Singla · 3 months 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.Log in to reply

I think E would also exist if angleBOA+angleBOC=90 – Sarthak Singla · 3 months ago

Log in to reply

– Deeparaj Bhat · 3 months ago

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