Question

ABCD is a rectangle. E is the midpoint of AB. Prove that ΔDEC is an isosceles triangle. (HINT : Prove using SAS, ΔAED = ΔBEC ).
And plz give answer step-by-step.​

Answer 1

Given :

Given that, ABCD is a rectangle. E is the midpoint of AB.

To prove :

We'll have to prove that ∆DEC is an isosceles triangle.

Proof :

Given that, E is the midpoint of the side AB.

In ∆AED and ∆BEC :

→ ∠DAE = ∠CBE

All the the angles in a rectangle are 90°.

→ AE = BE

E is the midpoint of AB.

→ AD = CB

Opposite sides in a rectangle are equal.

By SAS congruence rules, ∆AED ≌ ∆BEC.

→ ED = EC

By CPCT rule.

We know that, in a triangle if any two sides are congruent, then it is an isosceles triangle.

→ ∆DEC is an isosceles triangle.

Henceforth, proved!

Answer 2

Step-by-step explanation:

In triangle ADE and BCE

AD=BC(opposite sides of rectangle)

AE=BE (E is the midpoint of AB)

angles DAE and CBE are 90 degree

(all angles in rectangle are 90)

therefore ,

triangle DAE and CBE are congruent

(SAS congruence rule)

DE=CE ( by Cpct)

Two sides are equal So,triangle DEC is isosceles