Results 1 to 2 of 2

Thread: Triangle

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    38

    Triangle

    Given a $\displaystyle \triangle ABC$ and $\displaystyle D$ a point on the extension of side $\displaystyle BC$ (from $\displaystyle B$) such that $\displaystyle \overline{BD}=\overline{BA}$ and let $\displaystyle M$ be the mid point of $\displaystyle \overline{AC}$. The angle bisector of $\displaystyle \measuredangle\,ABC$ intersects to $\displaystyle \overline{DM}$ in $\displaystyle P$. Prove that $\displaystyle \measuredangle\,BAP=\measuredangle\,ACB.$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    • Let $\displaystyle N$ be the foot of the angle bisector of $\displaystyle B$ on the side $\displaystyle AC$. Since $\displaystyle \overline{AD}\parallel\overline{BN}\implies\triang le ADM\sim\triangle NPM$, it yields $\displaystyle \frac{{\overline {AD} }}
      {{\overline {NP} }} = \frac{{\overline {AM} }}
      {{\overline {MN} }}$
    • On the other hand, $\displaystyle \triangle BCN\sim\triangle DCA$, then
    $\displaystyle \frac{{\overline {BN} }}
    {{\overline {AD} }} = \frac{{\overline {CN} }}
    {{2 \cdot \overline {AM} }} \implies \frac{{2 \cdot \overline {MN} }}
    {{\overline {NP} }} = \frac{{\overline {CN} }}
    {{\overline {BN} }} \implies \frac{{\overline {AN} }}
    {{\overline {CN} }} = \frac{{\overline {BP} }}
    {{\overline {BN} }}$
    • Since $\displaystyle \frac{{\overline {AN} }}
      {{\overline {CN} }} = \frac{{\overline {AB} }}
      {{\overline {BC} }} \implies \frac{{\overline {BP} }}
      {{\overline {BN} }} = \frac{{\overline {AB} }}
      {{\overline {BC} }}$
    We have $\displaystyle \measuredangle\,ABN=\measuredangle\,CBN\implies\tr iangle{ABP}\sim\triangle{CBN}\,\therefore\,\measur edangle\,BAP=\measuredangle\,ACB$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 23rd 2011, 08:10 AM
  2. Replies: 3
    Last Post: Apr 30th 2009, 07:41 AM
  3. Replies: 1
    Last Post: Oct 28th 2008, 07:02 PM
  4. Replies: 7
    Last Post: Jul 19th 2008, 06:53 AM
  5. Replies: 27
    Last Post: Apr 27th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum