Geometry Problem. Post your solution in the comment box below.

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

Labels:
congruence,
double angle,
geometry problem,
isosceles,
quadrilateral

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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

https://photos.app.goo.gl/abtf6M4UGBGchacD7

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.

ReplyDelete