In figure,AB is perpendicular to AE,BC is perpendicular to AB,BE=DE and angle AED=120, find: (a) EDC (b)DEC (c)Hence prove that EDC is an eq. triangle?

Dear Student,

Please find below the solution to the asked query:

Given : AED  =  120° , AB is perpendicular on AE , So  BAE  =  90° and BC perpendicular to AB , So  ABD  =  90° 

From angle sum property of quadrilateral we get in quadrilateral ABDE :

BAE + AED + EDB + ABD = 360° , Substitute given values we get

90° + 120°  + EDB + 90° = 360° ,

300° + EDB  = 360° ,

EDB  = 60°

And

EDB  = EDC  = 60°                                                      ( Same angle )   (  Ans )

Also given DE  = CE so from base angle theorem we get

EDC  =  ECD =  60°

And from angle sum property of triangle we get in triangle EDC ,

EDC  + ECD  + DEC  = 180° ,

60° + 60° + DEC  = 180° ,
 
120° + DEC  = 180° ,

DEC  = 60°                                                                ( Ans ) 

Here we can see that all angles in triangle EDC are at 60° , SO we can say that triangle EDC is a equilateral triangle .   ( Hence proved )

Hope this information will clear your doubts about topic.

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RegardsE

  • 1
I am having a brilliant solution and you are not able to even draw the correct diagram but still acc. to your ques I'm making a correct diagram and here is the solution:

Angle A + Angle B + Angle BDE + Angle AED = 360 (ABDE is a quadrilateral),
Hence, 90 + 90 + Angle BDE + 120 = 360,

Angle BDE = 360 - (90+90+120),

Angle BDE = 60 ans.

Angle BDE = Angle ECD = 60 (angle opposite to equal sides are equal)

In ?ECD,
Angle CDE + Angle ECD + AngleCED = 180,

60+60+Angle CED=180,
Angle CED = 180 - (60+60)
Angle CED = 60.

  • 3
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