Question

In the given figure, ABCD is a parallelogram. P is the mid-point of AB. Prove that ACBM is a parallelogram.

Answer 1

(i) $ABCD$ is parallellogram

(ii) $P$ is midpoint pf $CM$.

$ABCD$ is parfallellogram

$\therefore AD||BC$ .

$\angle MAB=\angle CBA$ ,,,(1) (interior opposite angle of tranversal AB)

$\angle CPB=\angle ABM$ ...(2) (vertical opposite angle)

In $\Delta CPB$ and $\Delta ABM$

from (1) and (2) $\angle BCP=\angle PMA$

$CP+PM=\frac{CM}{2}$ ...(4) (P is mid point )

from (2), (3) and (4)

$\Delta CPB\cong \Delta MPA$

$\therefore CB=AM$

We know $CB$ is parallel to $AM$

$\therefore \angle AMB\text{is a parallelllogram since}CB||AMandCB=AM$