## Saturday, September 22, 2018

### Geometry Problem 1387: Quadrilateral, Double Angle, Congruence, Isosceles Triangle

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Extend AC to F such that DC = CF = d

< 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

1. Hence BE = EF and so b = c + d

2. 2nd Solution - more lengthy

Extend 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

3. Let the perpendicular bisector of AB intersect BD at F; easily ACDF is an isosceles trapezoid and EF=EC, CD=AF=BF, done.

Best regards

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

Draw 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

5. 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.