Proposed Problem
Click the figure below to see the complete problem 383 about Concave Quadrilateral, Angle Bisectors, Semi-difference of Angles.
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Complete Geometry Problem 383
Level: High School, SAT Prep
Tuesday, November 3, 2009
Problem 383. Concave Quadrilateral, Angle Bisectors, Semi-difference of Angles
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Extend DC to met AE in F. Ext. Ang.DFE = A/2 + β and C/2 = A/2 + β + x. Also, A/2 + α = x + C/2 or A/2 + α = x + A/2+ β + x which gives us:
ReplyDelete2x = α -β or x = (α -β)/2
Ajit
Let alpha = a and beta = b,
ReplyDeleteangle BAF=angle DAF=z,angle BCD=2y
Let seg. AE intersection seg.BC be point F.
In tri ABF,
angle ABF + angle FAB + angle AFB = 180
angle AFB =180-(z+a)
angle AFB = angle CFE...(vertex opposite angles)
angle CFE = 180-(z+a)...(1)
Quadrilateral ABCD is concave ,
hence,2y=a+2z+b
angle FCE = y=(a+2z+b)/2...(2)
By (1) and (2) in tri CFE,
x=180-[y+{180-(z+a)}]
x=180-[{(a+2z+b)/2} + {180-(z+a)}]
X=180-[{(a+2z+b-2z-2a)/2}+180]
x=180-[{(b-a)/2}+180]
x=180-[(b-a)/2]-180
x=(a-b)/2
[For easy typing, I use a for alpha & b for beta]
ReplyDeleteLet the intersection pt. of AE & BC be F
Consider triangle BAF
<BFE=a+<BAD/2 (ext. < of triangle)
<EFC=180-<BFE (adj. <s on st. line)
<EFC=180-a-<BAD/2
Consider triangle EFC
x=180-<EFC-<BCD/2
=180-(180-a-<BAD/2)-<BCD/2
=a+<BAD/2-<BCD/2----------(1)
Consider quad. AFCD
b=360-<FAD-reflex<BCD-<AFC
=360-<BAD/2-(360-<BCD)-(a+<BAD/2)
=<BCD-<BAD-a
So, a+b=<BCD-<BAD----------(2)
Sub (2) in (1)
x=a-(a+b)/2
=(a-b)/2