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Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, September 22, 2018
Geometry Problem 1387: Quadrilateral, Double Angle, Congruence, Isosceles Triangle
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Extend AC to F such that DC = CF = d
ReplyDelete< CFD = alpha < ABD and hence ABFD is a cyclic quadrilateral.
Since AE = ED ABFD is moreover a isoceles trapezoid.
Hence BC = CF and so b = c+d
Sumith Peiris
Moratuwa
Sri Lanka
Last line should read
DeleteHence BE = EF and so b = c + d
2nd Solution - more lengthy
ReplyDeleteExtend CA to X such that BA bisects < CBX
XBCD is concyclic and Tr.s XBC and AEC are similar
Using the above we can write 3 equations using b,c,d and AX,BX and AE which yield b = c+d
Sumith Peiris
Moratuwa
Sri Lanka
Let the perpendicular bisector of AB intersect BD at F; easily ACDF is an isosceles trapezoid and EF=EC, CD=AF=BF, done.
ReplyDeleteBest regards
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ReplyDeleteDraw circumcircle of triangle ABD ( O is the center)
OEM is the perpendicular bisector of AD
Let AE meet circle ABD at F
We have ^(AFD)=^(ABD) = alpha => ^(CDF)= alpha ( ^(ECD) is the external angle of triangle CDF)
And CDF is isosceles triangle => CF=CD=d
Due to symmetry with OEM as axis of symmetry
EB=EF= b= EC+CF= c+d
Parallel through C to AD. Let F be the intersection with BD. AFCD is isosceles trapezoid. Because of double angle AFB is going to be isosceles. qed.
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