The ΔBAE is congruent to ΔDAE.
Given to us,
E is the midpoint of ,
AC⊥ BD,
As given E is the midpoint of the line [tex]\overline{BD}[/tex],
So, BE = ED,
further, line AE = AE, because this is the common line between the two triangles.
Also, ∠BED =∠DEA =90°, as AC⊥ BD.
Therefore, the ΔBAE is congruent to ΔDAE.
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