Geometry Problem

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

Click the figure below to see the problem 740 details.

## Friday, April 13, 2012

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

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

ReplyDeletefrom 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

in triangle ABC: <A+<B+<C=180

ReplyDeletein triangle ABD: <ABD= 180-<A-<D

=A+B+C-A-(B+C)/2

=(B+C)/2

ABD is isoscele;AB=AD

Draw AE ⊥ BD

ReplyDelete∠EAD = 90 - (∠B + ∠C)/2 = ∠A/2

So AE bisects ∠A

Follows right ∆s AEB, AED are ≡

Hence AB = AD

<ADB=1/2(<ABC+<ACB)=1/2 (180-<BAC)=90-1/2 <BAC

ReplyDeleteBisect <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

the solution is uploaded to the following link:

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