BdMO National Junior 2020 P8

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BdMO National Junior 2020 P8

Unread post by Mursalin » Thu Feb 04, 2021 12:28 am

\(ABC\) ত্রিভুজে \(\angle B = 50^\circ\) এবং \(\angle C = 60^\circ\)। \(D, BC\)-এর মধ্যবিন্দু। ত্রিভুজ \(ABC\)-এর পরিবৃত্ত হল এমন একটি বৃত্ত যা ত্রিভুজের তিনটি শীর্ষবিন্দু দিয়ে যায়। \(ACD\) এবং \(ABD\)-এর পরিবৃত্ত \(AB\) এবং \(AC\)-কে যথাক্রমে \(F\) এবং \(E\) বিন্দুতে ছেদ করে। \(AEF\)-এর পরিবৃত্তের কেন্দ্র \(O\)। \(\angle FDO=\)?

Let \(ABC\) be a triangle where \(\angle B = 50^\circ\) and \(\angle C = 60^\circ\). \(D\) is the midpoint of \(BC\). The circumcircle of a triangle \(ABC\) is defined to be the circle going through the three vertices. The circumcircles of \(ACD\) and \(ABD\) intersects \(AB\) and \(AC\) at \(F\) and \(E\) respectively. The circumcentre of \(\triangle AEF\) is \(O\). \(\angle FDO=\)?
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Re: BdMO National Junior 2020 P8

Unread post by Enthurelxyz » Tue Feb 09, 2021 8:02 pm

As, $O$ is the center of $\triangle AFE$ and $\angle FAE=70$ so $\angle FOE=140$ so $ \angle OEF=20$.

As, $AFDC$ is cyclic then $\angle FDB=70$. As, $AEDB$ is cyclic, it is easy to prove that $\angle EDC=70$. So, $\angle FDE=40$.

$\angle FDE+\angle FOE=180$ so $OFDE$ is cyclic so $\angle OEF=\angle ODF=20$.
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