# Solve this : Plz don't spam

Dear Student,

In a parallelogram opposite angles are equal therefore,
Now, ∠∠ABC + ∠∠EBC = 180 (Linear Pair angles)
∠∠EBC = 180 - ​ ∠∠ABC ............. (1)

Also, ​∠∠ADC + ∠∠CDF = 180 ​(Linear Pair angles)
∠∠CDF = 180 - ​∠∠ADC ............... (2)
From (1) and (2) ,
∠∠EBC = ​∠∠CDF ..............(3) (Since ​∠∠ABC = ∠∠ADC )

Opposites sides of a parallelogram are equal therefore
AB = DC
but, AB = BE
hence, DC = BE ............ (4)

Similarly, DF = BC ............ (5)

In triangle BEC and triangle DCF

BE = DC (From (4) )
∠∠EBC = ​∠∠CDF (From (3) )
BC = DF (From (5) )

Therefore the two triangles are similar by SAS (Side Angle Side) test

Regards

• 1
Dear Student,

In a parallelogram opposite angles are equal therefore,
Now, ∠∠ABC + ∠∠EBC = 180              (Linear Pair angles)
∠∠EBC = 180 - ​ ∠∠ABC   ............. (1)

Also, ​∠∠ADC + ∠∠CDF = 180              ​(Linear Pair angles)
∠∠CDF = 180 - ​∠∠ADC   ............... (2)
From (1) and (2) ,
∠∠EBC = ​∠∠CDF      ..............(3)       (Since ​∠∠ABC = ∠∠ADC )

Opposites sides of a parallelogram are equal therefore
AB = DC
but, AB = BE
hence, DC = BE        ............ (4)

Similarly, DF = BC  ............ (5)

In triangle BEC and triangle DCF

BE = DC                      (From (4) )
∠∠EBC = ​∠∠CDF          (From (3) )
BC = DF                      (From (5) )

Therefore the two triangles are similar by SAS (Side Angle Side) test

Regards
• 1

## ∠BAD=∠CDF (Corresponding angles for parallel lines AB and CD) ∠BAD=∠CBE (Corresponding angles for parallel lines AB and CD) Thus, ∠CDF=∠CBE (I) We know, AD=BC (ABCD is a parallelogram) and AD=DF (Given) Thus, DF=BC (II) Similarly, BE=CD (III) Now, In △CDF and △CBE ∠CDF=∠CBE (From I) FD=BC (From II)  BE=CD (From III) Thus, △FDC≅△CBE (SAS rule) △BEC≅△DCF

• 1
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