Question

Prove that opposite sides and angles of a parallelogram are equal

Answer 1

Consider a ||gm ABCD, such that AB||CD and BC||AD.
Now, consider $\Delta ABD\&\Delta BCD$,
$\angle BDC=\angle ABD(Alt.int.\angle S)$ ------------ (1)
$\&\angle ADB=\angle DBC(Alt.int.\angle S)$ ---------- (2)
BD = BD (Common)
So, by ASA congromy, $\Delta ABD\cong \Delta BCD$
So, AD = BC and AB = DC (CPCT)
Also, (1) +(2) gives, $\angle ADC=\angle ABC$
||ly you can prove, $\angle DAB=\angle DCB$