The correct option is
A True
Given ABCD is a ∥gm & AM=NC
So AB∥ DC, AB=DC, AD∥ BC, AD=BC
∠BAC=∠ACD [ Alternate Angles ]
∠CAD=∠BCA [ Alternate Angles ]
In △ AMB & △ CND,
AB=DC
∠MAB=∠NCD[ Alternate Angles ]
AM=NC
So △ AMB≅△ CND
So ∠CND=∠AMB,∠NDC=∠MBA
MB=ND ---(1)
Now ∠DNA=∠BMC{180o−∠CND=180o−∠AMB⇒Straight Angles}
So MB∥ ND ---(2)
[ Since ∠DNA=∠BMC=Alternate Angles if MN is transversal for MB & ND ]
Similarly taking △ AMD & △ CNB, we can prove them to be congruent.
So ∠CNB=∠AMD,∠NBC=∠MDA
MD=BN ---(3)
Now ∠DMC=∠BNA{180o−∠AMD=180o−∠CNB⇒Straight Angles}
So MD∥ BN ---(4)[ Since ∠DMC=∠BNA=Alternate Angles if MN is transversal for MD & BN ]
From (1),(2),(3)&(4) , we get that BMDN is a parallelogram.
![363360_183036_ans.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/363360_183036_ans.png)