Quadrilaterals — Maths STD 9 — Question

$\angle\text{ABM}=\angle\text{CBM}\ ...(1) (BM$ bisects $\angle\text{B})$
$\angle\text{ABM}=\angle\text{MDC}\ ...(2)($Alternate angles$)$
$\angle\text{CBM}=\angle\text{ADM}\ ...(3)($Alternate angles$)$
From equations $(1), (2)$ & $(3)$
$\angle\text{MDC}=\angle\text{ADM}...(4)$
Now, consider $\triangle\text{ABM}$ & $\triangle\text{CBD}$
$\angle\text{CBD}=\angle\text{ABD} \{$from eq $(1)\}$
$DB = DB ($Common$)$
$\angle\text{ADB}=\angle\text{CDB} \{$from eq $(4)\}$
Hence, by ASA property,
$\triangle\text{ADB}\cong\triangle\text{CBD}$
$⇒ AB = CB, AD = CD$
Hence, it becomes a Rhombus.
So now diagonals of a Rhombus bisect each other at $90^\circ .$
$\Rightarrow\angle\text{AMB}=90^\circ$