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

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, July 28, 2018

### Geometry Problem 1370: Equilic Quadrilateral, 120 Degrees, Congruence, Midpoint, Diagonal, Equilateral Triangle

Labels:
120,
angle,
equilateral,
equilic quadrilateral,
geometry problem

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Extend AB & CD to meet at H, where m(H)=60=m(BJC)

ReplyDeleteBHJC is concyclic

let m(HBJ)=m(HCJ)=x

=> m(ABJ)=180-x=m(DCJ)

Hence ABJ and DCJ are congruent => AJ=JD -------(1)

Let m(BAJ)=m(CDJ)=y

Since AJ=DJ, in the triangle AJD m(JAD)=m(JDA)

=> @-y=120-@+y (@ to be treated as Alpha)

=> @-y=60=m(JAD) --------(2)

From (1) and (2) AJD is equilateral

Triangles ABJ & CDJ are congruent SAS and the result follows

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Extend AB and DC to P. Ang PBJ = ang PCJ => tr ABJ congr to tr DCJ

ReplyDelete=> a) AJ=DJ and ang BJA = ang DJC => b) Ang AJD = 60°

From a) and b) AJD equilateral