Question

Prove that in a regular pentagon, the perpendicular from any vertex to the opposite side bisects that side.

Answer 1

Consider the regular pentagon, ABCDE:

Construction: Draw DF perpendicular to AB. Join AD and DB.

In ΔAFD and ΔBFD:

AD = BD (Diagonals of a regular pentagon are equal in length)

DF = DF (Common)

∠AFD = ∠BFD = 90°

∴ ΔAFD ≅ ΔBFD (RHS congruence criterion)

⇒ AF = FB (By c.p.c.t.)

Thus, in a regular pentagon, the perpendicular from any vertex to the opposite side bisects that side.