Exercice AA-004-G

  • Math&Maroc

    ABCABC triangle tel que BAC=120\angle {BAC}=120^\circ . On suppose que les bissectrices des angles A,B\angle {A}, \angle {B} et C\angle {C} coupent les côtés opposés aux points D,ED,E et FF respectivement.
    Montrer que le cercle de diamètre [EF][EF] passe par le point DD.

    Wenn ich nur erst die sätze habe ! die beweise werde ich schon finden (Bernhard Riemann).

  • Math&Maroc

    Geometry enlightens the intellect and sets one's mind right.
    All its proofs are very clear and orderly.
    It is hardly possible for errors to enter into geometrical reasoning, because it is well arranged and orderly.
    Thus, the mind that constantly applies itself to geometry is not likely to fall into error.
    In this convenient way : the person who knows geometry acquires intelligence.

    Ibn Khaldoun ( May 27, 1332 - March 19, 1406), Al Muqaddimah.

    Short biography : Ibn Khaldun was a North African polymath — an astronomer, economist, historian, Islamic scholar, Islamic theologian, hafiz, jurist, lawyer, mathematician, military strategist, nutritionist, philosopher, social scientist and statesman—born in North Africa in present-day Tunisia.

  • On applique le loi de sinus dans le triangle BAD on obtient :
    sin(60)sin(B)\frac{sin(60)}{sin(B)}= BDAD\frac{BD}{AD} <=> 32sin(B)\frac{\sqrt3}{2sin(B)}= BDAD\frac{BD}{AD}
    Et on On applique le lois de sinus dans le triangle BAC on obtient :
    sin(120)sin(B)\frac{sin(120)}{sin(B)}= BCAC\frac{BC}{AC} <=> 32sin(B)\frac{\sqrt3}{2sin(B)}= BCAC\frac{BC}{AC}
    et puisque bisecteur de C coupe AB dans F alors BCAC\frac{BC}{AC}= BFAF\frac{BF}{AF}
    donc : BDAD\frac{BD}{AD}=BFAF\frac{BF}{AF} alors ∠ADF=∠FDB
    de meme on montre que : ∠ADE=∠EDC
    Donc : ∠EDF=90 => EF diametre du cercle passant par D,EetF .

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