Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ , and let $O$ , $H$ be the triangle's circumcenter and orthocenter respectively . Let also $A^{'}$ be the point where the angle bisector of the angle $BAC$ meets $\omega$ . If $A^{'}H=AH$ , then find the measure of the angle $BAC$.
Problem
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Tags: geometry, circumcircle, rhombus, angle bisector
johnkwon0328
04.02.2015 17:48
$\angle A=\frac{\pi}{3}$
Submathematics
05.02.2015 19:22
We know $AH=2RCosA$.............(1) $OA=OA'$ and $AH=HA'$ AND $AH$ PARALLEL TO $OA'$ give $AHA'O$ is a rhombus, implying $R=AH$. Putting in (1). we get $A=60$
combidude
05.10.2015 22:56
johnkwon0328 wrote: $\angle A=\frac{\pi}{3}$ What is your logic/solution?
OlympusHero
28.07.2021 01:32
Can I get a hint on this question please, thanks.
LeYohan
19.12.2024 17:10
$\angle BAH = \angle OAC$ and because $AA'$ is the angle bisector of $\angle BAC \implies \angle HAA' = \angle OAA'$. Notice that both $\triangle AHA'$ and $\triangle AOA'$ are isosceles which means they are congruent. This means that $AH = OH$ and it's well known that this is true $ \iff \angle BAC = 60^{\circ}$.