Answer this question

Answer:
So [tex]\angle ABC\cong \angle ADC[/tex]
And [tex]\angle BAD\cong \angle BCD[/tex]
Step-by-step explanation:
Given;
ABCD is a parallelogram and opposite sides of this parallelogram are parallel that is 'AB' parallel with 'DC' and 'BC' parallel 'AD'
Now joint the point 'A' and 'C' and we get two different triangle 'ABC' and 'ADC',
Reason: Given [tex]BC\parallel AD[/tex] then mention that Alternate Interior angles are equal for parallel lines.
Reason: Given [tex]BC\parallel AD[/tex] then mention that Alternate Interior angles are equal for parallel lines.
Reason: Common side.
Then [tex]\bigtriangleup ACB\cong \bigtriangleup ACD[/tex]
∴ [tex]\angle ABC\cong \angle ADC[/tex]
By using diagonal 'BD' we could flow similar argument to prove that [tex]\bigtriangleup BAD\cong \bigtriangleup BCD[/tex] and also [tex]\angle BAD\cong \angle BCD[/tex]
∴ [tex]\angle BAD\cong \angle BCD[/tex]