ABCDE is a regular pentagon. Let’s prove ABC is an isosceles triangle and BE and CD are two parallel line segments.

For first part of the proof let us draw a figure.

ABCDE is a regular pentagon.

To Prove: Δ ABC is an isosceles triangle

Proof:

In the question it is mentioned that the given pentagon is a regular pentagon hence all the sides of this pentagon has equal lengths.

So in Δ ABC,

Length of AB = length of BC [as they both are sides of the regular pentagon]

Now for the given triangle, side AB = BC,

So ∠ BAC = ∠ ACB ………… (Angles opposite to the equal sides in a triangle are equal. One of the important property of isosceles triangle]

Hence it is proved that the given Δ ABC is an isosceles triangle.

Now for the second proof let us draw a diagram again.

To Prove: BE || CD

Proof:

In the question it is mentioned that the given pentagon is a regular pentagon hence all the sides of this pentagon has equal lengths.

So in Δ ABE,

Length of AB = length of AE [as they both are sides of the regular pentagon]

Now for the given triangle, side AB = AE,

So ∠ ABE = ∠ AEB ………… (Angles opposite to the equal sides in a triangle are equal. One of the important property of isosceles triangle]

Now we also know that BC = DE

Also ∠ ABC = ∠ AED ………. [For a regular pentagon all interior angles are equal]

Now we know that ∠ ABC = ∠ ABE + ∠ EBC …………. (i)

Similarly for ∠ AED = ∠ AEB + ∠ BED …………………… (ii)

Now subtracting equation (ii) from (i)

∠ ABC - ∠ AED = (∠ ABE + ∠ EBC) – (∠ AEB + ∠ BED) … [∠ ABC = ∠ AED]

0 = ∠ ABE + ∠ EBC - ∠ AEB - ∠ BED …… [∠ ABE = ∠ AEB]

∠ EBC - ∠ BED = 0

∠ EBC = ∠ BED

∠ BCD = ∠ CDE ………. [For a regular pentagon all interior angles are equal]

Now from the figure it is observed that BCDE is a quadrilateral.

For the given quadrilateral all its interior angles are equal and hence for the angles to be equal BE has to be parallel to CD.

Hence Proved.