Friday, April 13, 2012

Problem 740: Scalene Triangle, Angle, Semi-sum, Cevian, Isosceles Triangle

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 740 details.

Online Geometry Problem 740: Scalene Triangle, Angle, Semi-sum, Cevian, Isosceles Triangle.

5 comments:

  1. http://img259.imageshack.us/img259/2267/problem740.png

    from C draw line CE ( E on AB extend) such that ∠ACE= ½(∠B+∠C) ( see sketch)
    in triangle ACE , ∠AEC= 180- ∠A-1/2(∠B+∠C) =1/2(∠B+∠C) since ∠A+∠B+∠C=180
    So ∆ACE is isosceles .
    ∠ADB=∠ACE => BD // EC => ∆ABD is isosceles and AB=AD

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  2. in triangle ABC: <A+<B+<C=180
    in triangle ABD: <ABD= 180-<A-<D
    =A+B+C-A-(B+C)/2
    =(B+C)/2
    ABD is isoscele;AB=AD

    ReplyDelete
  3. Draw AE ⊥ BD
    ∠EAD = 90 - (∠B + ∠C)/2 = ∠A/2
    So AE bisects ∠A
    Follows right ∆s AEB, AED are ≡
    Hence AB = AD

    ReplyDelete
  4. <ADB=1/2(<ABC+<ACB)=1/2 (180-<BAC)=90-1/2 <BAC
    Bisect <BAC with AE WHICH MEETS BD AT E. THEN <AED=180-<A/2-(90-<A/2)=90
    SO, TRIANGLES AEB AND AED ARE CONGRUENT SINCE TWO ANGLES 90, A/2 ARE EQUAL AND AE IS COMMON (AAS)
    SO AB=AD

    ReplyDelete
  5. the solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJN3Q0elRfVkM5WjA

    ReplyDelete