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Math Help - Triangle

  1. #1
    Junior Member
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    Triangle

    Given a \triangle ABC and D a point on the extension of side BC (from B) such that \overline{BD}=\overline{BA} and let M be the mid point of \overline{AC}. The angle bisector of \measuredangle\,ABC intersects to \overline{DM} in P. Prove that \measuredangle\,BAP=\measuredangle\,ACB.
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  2. #2
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    • Let N be the foot of the angle bisector of B on the side AC. Since \overline{AD}\parallel\overline{BN}\implies\triang  le ADM\sim\triangle NPM, it yields \frac{{\overline {AD} }}<br />
{{\overline {NP} }} = \frac{{\overline {AM} }}<br />
{{\overline {MN} }}
    • On the other hand, \triangle BCN\sim\triangle DCA, then
    \frac{{\overline {BN} }}<br />
{{\overline {AD} }} = \frac{{\overline {CN} }}<br />
{{2 \cdot \overline {AM} }} \implies \frac{{2 \cdot \overline {MN} }}<br />
{{\overline {NP} }} = \frac{{\overline {CN} }}<br />
{{\overline {BN} }} \implies \frac{{\overline {AN} }}<br />
{{\overline {CN} }} = \frac{{\overline {BP} }}<br />
{{\overline {BN} }}
    • Since \frac{{\overline {AN} }}<br />
{{\overline {CN} }} = \frac{{\overline {AB} }}<br />
{{\overline {BC} }} \implies \frac{{\overline {BP} }}<br />
{{\overline {BN} }} = \frac{{\overline {AB} }}<br />
{{\overline {BC} }}
    We have \measuredangle\,ABN=\measuredangle\,CBN\implies\tr  iangle{ABP}\sim\triangle{CBN}\,\therefore\,\measur  edangle\,BAP=\measuredangle\,ACB
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