Geometry Problem
Click the figure below to see the complete problem 507 about Triangle, Interior Point, 120 Degrees, Congruence, Angle.
See also:
Complete Problem 507
Level: High School, SAT Prep, College geometry
Friday, August 20, 2010
Problem 507: Triangle, Interior Point, 120 Degrees, Congruence, Angle
See also:
Complete Problem 507
Level: High School, SAT Prep, College geometry
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x=30
ReplyDeleteDraw CG = AC, G on extension of AD
ReplyDelete=> ▲DGC isoceles, => ▲BGD equilateral
=> B, D, C are on circle with center G
=> x = 1/2 arc BD, BGD = arc BD, => x = 1/2 BGD
=> x = 30°
from D extend line (DK) such as BD=DK=BK and A,D,and K are collinear
ReplyDeleteand from A extend line (AH) such as A,C and H being collinear points and AK = AH
for notation ,let BD=m
we conclude that AD = CH
and for notation again let AD=n
now , let P be a point on (KH) such that DC=KP
so we have traingles ▲BKP and ▲BDC are simillar
and therfore we conclude thatt ▲BCP is isocelles
also we have (DK)//(CP) and then CP=CH
let ∠CBP=2θ we conclude easily that x=a+θ
as a result we have ∠DCP=90°+x-θ=90+a=∠ADC and CP=AD=n
thus DP=m and so far DPCA is a parallelogram
focus now in isocelles triangles ▲BDP and ▲DKP we have
∠DKP=∠DPK=90°-θ
therfore ∠DPB=90°-a-∠BPK=90°-a-x
then in ▲DKP we have 60+2a +2(90°-a-x)=180°
thus x=30°