In the given figure, D is the midpoint of BC, DE ⊥ AB and DF ⊥ AC such that DE=DF. Then, which of the following is true?
From the given figure, we have
D is the mid-point of BC
Also, DE is perpendicular to AB
DF is perpendicular to AC
And, DE = DF
Now, in ∆BED and ∆CFD we have:
DE = DF
BD = CD
∠E = ∠F = 90o
∴ By RHS congruence rule
∆BED ≅ ∆CFD
Thus, ∠B = ∠C
AC = AB
Hence, option (A) is correct