Inspired by one of the latest numberphile videos I started playing around with the specific configuration.

I would like to prove the following:

Given is a random $\triangle ABC$ with perpendicular bisector $MS$ and bisector $AS$. (see figure)

If $\widehat{SBA} > 90°$ then $\widehat{ACS} < 90°$

unknown theorem

Experimenting in Geogebra gave me some clues. If I could prove $\widehat{BAS} = \widehat{SBM}$ (which is quite surprising) then proving the conjecture above would be a breeze.

However this task seems quite challenging, I tried proving $\triangle BMS \sim \triangle AHS$ without success. Could someone give me some pointers?

(Is this a known theorem?)


Let $O$ be the circumcenter of the $\triangle ABC$. Then, the $\angle SOC = \angle MOC = \angle BAC$.
Also, $\angle SAC = \frac12 \angle BAC$.

Thus, $O$ is also the circumcenter of the $\triangle SAC$ (because the angle subtended by $SC$ at $O$ is twice the angle subtended by $SC$ at $A$).

Thus, the points $A,B,S$ and $C$ all lie on the same circle. Therefore $ABSC$ is a cyclic quadrilateral. Thus, if angle $SBA > 90^\circ$, then $\angle ACS$ is $< 90^\circ$.

  • $\begingroup$ That's a nice way to prove the conjecture. Cool! Do you know a way to prove $\widehat{BAS}=\widehat{SBM}$? $\endgroup$ – dietervdf Nov 13 '14 at 20:18
  • 1
    $\begingroup$ Once we have proved that the quadrilateral is cyclic, it's straightforward. $\widehat{BAS}=\widehat{SAC}=\widehat{SBM}$. $\endgroup$ – Ojas Nov 13 '14 at 20:29

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