Question

4. In figure AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°, then find ∠ECD ​

Answer 1

Answer:

Given In figure AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°, then find ∠ECD.

Given angle AED = 120 degree.

AB is perpendicular to AE

Therefore angle EAB = 90 degree

BC is perpendicular to AB

Therefore angle DBA = 90 degree

From angle sum property of quadrilateral ABDE we get

Angle EAB + angle AED + angle EDB + angle DBA = 360 degree

So 90 + 120 + angle EDB + 90 = 360 degree

300 + angle EDB = 360

Or angle EDB = 360 – 300

Or angle EDB = 60 degree

So angle ECD = 60 degree (since similar angle)

Also CE = DE

From base angle theorem we get

Angle EDC = angle ECD = 60 degree

Answer is 60 degree

hope it's help full Army

it's wrong I am so sorry

Answer 2

Given: In figure AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°.

To find: The measure of ∠ECD.

Solution:

As evident from the question, the triangle ∠CDE is isosceles because CE=DE. Thus, angles ∠C and ∠D are equal. Let the angles ∠ECD and ∠EDC be x each and let the angle ∠DEC be y. Since the sum of the angles of a triangle is equal to 180, the following equation can be written.

$x + x + y = 180$

The angles ∠ECD and ∠CEA are internal alternate angles and hence, they are equal. The angle ∠CEA can be written as follows.

$\angle CEA = 120 - y$

Now, the equation can be written as

$\angle ECD = \angle CEA$

$x = 120 - y$

$y = 120 - x$

The (y) can be substituted as (120-x) in the first equation.

$x + x + 120 - x = 180$

$x = 60$

Therefore, the measure of ∠ECD is 60°.