# 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,

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

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

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

90$°$ + 120$°$  + $\angle$ EDB + 90$°$ = 360$°$ ,

300$°$ + $\angle$ EDB  = 360$°$ ,

$\angle$ EDB  = 60$°$

And

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

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

$\angle$ EDC  =  $\angle$ ECD =  60$°$

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

$\angle$ EDC  + $\angle$ ECD  + $\angle$ DEC  = 180$°$ ,

60$°$ + 60$°$ + $\angle$ DEC  = 180$°$ ,

120$°$ + $\angle$ DEC  = 180$°$ ,

$\angle$ 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 )

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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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