hard

[dektep]

In acute \triangle{ABC} denote by M and N the midpoints of the altitudes BB_{1} and CC_{1} respectively,P = AM \cap CC_{1} and Q = AN \cap BB_{1}.If the points B,C,P and Q are concyclic then prove that \triangle{ABC} is isosceles
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[PDatK40SP]

Nguyên văn bởi dektep

In acute $\triangle{ABC} $ denote by $M $ and $N $ the midpoints of the altitudes $BB_{1} $ and $CC_{1} $ respectively,$P = AM \cap CC_{1} $ and $Q = AN \cap BB_{1} $.If the points $B,C,P $ and [TEX]Q /TEX] are concyclic then prove that $\triangle{ABC} $ is isosceles

$H $ is the orthocenter
$B,C,P,Q $ are concyclic means $\frac{ \bar{PC_1} }{\bar{PH} } = \frac{ \bar{QB_1} }{\bar{QH} } \ (1) $
We have:
$\frac{ \bar{PC_1} }{\bar{PH} }: \frac{ \bar{CC_1} }{\bar{CH} } =(C_1HPC) = (BHMB_1) = \frac{ \bar{MB} }{\bar{MH} }: \frac{ \bar{B_1B} }{\bar{B_1H} } = \frac{1}{2} \frac{ \bar{HB_1} }{ \bar{HM }} \ (2) $
And, similarly:
$\frac{ \bar{QB_1} }{\bar{QH} }: \frac{ \bar{BB_1} }{\bar{BH} } =(B_1HQB) = (CHNC_1) = \frac{ \bar{NC} }{\bar{NH} }: \frac{ \bar{C_1C} }{\bar{C_1H} } = \frac{1}{2} \frac{ \bar{HC_1} }{ \bar{HN }} \ (3) $
From $(1),(2),(3) $ we have:
$\frac{ \bar{BB_1} }{ \bar{BH} } : \frac{ \bar{CC_1} }{ \bar{CH} } = \frac{ \bar{HB_1} }{ \bar{HM} } : \frac{ \bar{HC_1} }{ \bar{HN} } \Rightarrow \frac{ \bar{BB_1} }{ \bar{CC_1} } = \frac{ \bar{HB} }{ \bar{HC} } . \frac{ \bar{HB_1} }{ \bar{HC_1} } : \frac{ \bar{HM} }{ \bar{HN} } = \frac{ \bar{HN} }{ \bar{HM} } $
And the equality $\frac{ \bar{BB_1} }{ \bar{CC_1} } = \frac{ \bar{HN} }{ \bar{HM} } $ leads us to $BB_1 =CC_1 $, as desired
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