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
Subscribe to:
Post Comments (Atom)
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
<ADB=(<ABC+<ACB)/2
ReplyDelete<ADB=(180-<BAC)/2
<ADB=90-<BAC/2
<BAC=180-2<ADB
<ABD=180-<BAD-<ADB
<ABD=180-(180-2<ADB)-<ADB=<ADB
As <ABD=<ADB, AB=AD (sides opp eq. <s)
Let < ABD = p and let < ADC = q
ReplyDeleteSo < DBC = q - C = B - p ==> q = B + C - p ......(1)
But 2q = B + C ......(2)
(2) - (1) gives q = p
Hence AB = AD
Sumith Peiris
Moratuwa
Sri Lanka