MCQ
$\text{ABCD}$ is a parallelogram, $M$ is the mid $-$ point of $BD$ and $BM$ bisects $\angle\text{B}.$ Then, $\angle\text{AMB}=$
- A$45^\circ$
- B$60^\circ$
- ✓$90^\circ$
- D$75^\circ$
✓
Answer
Correct option: C.
$90^\circ$
$\angle\text{ABM}=\angle\text{CBM}\ ...(1)$
$(BM$ 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 $\text{ASA}$ property,
$\triangle\text{ADB}\cong\triangle\text{CBD}$
$\Rightarrow AB = CB, AD = CD$
Hence, it becomes a Rhombus.
So now diagonals of a Rhombus bisect each other at $90^\circ .$
$\Rightarrow\angle\text{AMB}=90^\circ$
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