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Acute-angled triangle ABC is inscribed in circle $c$. In the smaller arc BC we choose a random point F and draw a line parallel to AC which intersects side AB in point Q. Then from point Q we draw a parallel to FC which intersects side AC to point H. From vertex B we draw a line parallel to FC which intersects the circle at point S. Show that angle SHC is equal to BAC.

I have tried the following: I have extended line SH until it meets the circle at point which I name G. Clearly G lies also on the line QF but we don't know it yet.

Since FC and BS are parallel chords of the circle, then BFCS must be an isosceles trapezoid, hence arcs BC and FS are equal. Therefore angle CAB is equal to angle SGF (inscribed angles of the same circle, having equal arcs). Then obviously SHE is equal to SGF.

But how can I prove that point G also lies on the parallel from F?

Or, is there any other way to prove that the required angles are equal?

Thank you in advance.

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Extend $FQ$ to meet the circle at point $L$.Let the points at which $FL$ and $CA$ intersect $BS$ be $T$ and $U$ respectively. Draw $AL$, $LS$ and $LH$$[$We haven't yet proved that points $L$,$H$ and $S$ are collinear.$]$

$AC\parallel FL$ $\Rightarrow$ Quadrilateral $ACFL$ is an isosceles trapezoid.

Since $QH\parallel CF$, quadrilateral $ALQH$ is an isosceles trapezoid as well.

$\Rightarrow$ $\angle ALH=\angle AQH=\angle ABU=\angle ABS=\angle ALS$

$\Rightarrow$ $\angle ALH=\angle ALS$

Hence, points $L$, $H$ and $S$ are collinear.

Thus, $\angle SHC=\angle SLF=\angle HLQ=\angle QAH=\angle BAC$

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