The proof of AM being congruent to CM is as follows:
Given that,
ABCD is the ║gm
Diagonals AC and BD with M as the intersection point.
To prove:
AM ≅ CM
Proof:
In ΔAMD and ΔBMC,
∠M = ∠M (∵ Common)
AD = BC (∵ opposite sides of ║gm are equal)
∠ADM = ∠CBM (∵ Alternate angle)
∵ ΔAMD ≅ ΔBMC
Thus, AM ≅ CM (using AAS property).
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