Question

In the adjoining figure, ABCD is a square and ABE is an equilateral triangle. Prove that ΔECD is an isosceles triangle.

Answer 1

Given:

• ABCD is a square.
• ABE is an equilateral triangle with AB as a side.

To Prove:

• Triangle ECD is isosceles i.e., EC = ED.

Proof [Step-wise]:

1. Let M be the midpoint of AB and N the midpoint of CD. 

In a square, the midpoint of AB is directly below the midpoint of CD.

So, M and N lie on the same line perpendicular to AB and CD.

2. The perpendicular bisector of AB is the line through M perpendicular to AB.

In an equilateral triangle ABE the vertex E lies on the perpendicular bisector of the base AB.

The altitude from E both bisects AB and is perpendicular to AB. 

Hence, E lies on the perpendicular bisector of AB. Use the standard fact that the altitude from the vertex of an equilateral triangle bisects the base.

3. Because AB and CD are parallel equal sides of the square and their midpoints M and N are collinear on the same perpendicular, the perpendicular bisector of AB coincides with the perpendicular bisector of CD, the same line through M and N is perpendicular to both AB and CD and bisects each.

4. From steps 2–3, E lies on the perpendicular bisector of CD.

A point on the perpendicular bisector of a segment is equidistant from the segment’s endpoints; therefore, EC = ED. 

This is the standard perpendicular-bisector property: each point on a perpendicular bisector is equidistant from the extremities of the segment.

5. EC = ED means triangle ECD has two equal sides, so it is isosceles.