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Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$

Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle of $ABC$ on $D$ and $E$ respectively . We suppose that $DE=BC$

What is the angle $\widehat{BAC}$

I tried using law of sines in triangle Also , let O be center of circle so OD=OE=r

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  • $\begingroup$ @user24142 we use thales when we have parallel lines ? Or not ! $\endgroup$
    – user233658
    Commented Dec 18, 2015 at 15:49
  • $\begingroup$ I see it now : so angles BAE=BCE=DBE=90 @user24142 $\endgroup$
    – user233658
    Commented Dec 18, 2015 at 15:55
  • $\begingroup$ @user24142 but what is the point? $\endgroup$
    – user233658
    Commented Dec 18, 2015 at 15:59
  • $\begingroup$ What does "passing by" mean? If it means "containing" or "intersecting" then the perpendicular to "AB" containing C is the line AC and there is no perpendicular to a AC containing B as AB is parallel to the perpendicular to AC. $\endgroup$
    – fleablood
    Commented Dec 18, 2015 at 18:00

3 Answers 3

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enter image description here

Note that $\angle BAC=\angle BEC$, and that $\angle BAC=180^\circ-\angle DFE=\angle CFE$. As $DE=BC$, $\angle BEC=\angle DCE$. Therefore, $\angle BAC = 60^\circ$.

Appendix:

enter image description here

See the comment section below.

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  • $\begingroup$ Can tou please explain the seconde line of your answer $\endgroup$
    – user233658
    Commented Dec 18, 2015 at 16:32
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    $\begingroup$ One easy way to do this is to see that $BFC$ and $DFE$ are congruent as $BC=DE$ and the two equal angles, and get $FC=FE=EC$. $\endgroup$
    – Sawarnik
    Commented Dec 18, 2015 at 20:35
  • $\begingroup$ @user233658 you may refer to the second figure. Since $DE=BC$, they have the same central angle, and, by halving it, the corresponding inscribed angle. But the problem is that each of the chords has two inscribed angles, and we don't know which is the congruent pair. Now let $\angle BEC=x$. Consider quadrilateral $DCEP$. We know either $\angle DCE=180^\circ-x$ (and $\angle DPE=x$), or $\angle DCE=x$ (and $\angle DPE=180^\circ -x$). If it is the former, then $\angle BEC+\angle DCE + \angle CFE=180^\circ+\angle CFE>180^\circ,$ which is absurd, so $\angle BEC=\angle DCE.$ $\endgroup$
    – James Pak
    Commented Dec 19, 2015 at 16:46
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A diagram of the setup

The above is basically what you're given. I've labeled a lot of the angles with $\alpha, \beta, \gamma, \delta$ and $\epsilon$. Two inscribed angles are equal when they're based off the same chord, ie I'm leaning heavily on the third diagram here.

Now note that $\alpha + \beta + \gamma + \delta + \epsilon = \pi$ and $\gamma + \delta = \gamma +\epsilon = \pi/2$. Furthermore, $\triangle BPC \cong \triangle DPE$ (angle side angle), so $\triangle BED \cong \triangle DCB$ (side side side) and we conclude $\gamma = \epsilon +\delta$. Therefore, $$\pi = (\gamma + \delta) + (\gamma + \epsilon) = 3\gamma$$ and so $\gamma = \pi/3$.

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Let $AECFBD $ (labeled anti-clockwise) be a regular hexagon inscribed in a circle center $O$. All the given conditions are satisfied by this assumption ( perpendicularities) and $(DE=BC).$ Simply by symmetry, angle BAC = $60^0.$

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