Question

point lies on the side bc of parallelogram ABCD such that BE =CE and BC = 2ab if angle ade = 50° then angle cde equals to​

Answer 1

Answer:

65

Step-by-step explanation:

Answer 2

Given:

Parallelogram ABCD.

E is the midpoint of side BC.

BC = 2AB

∠ADE = 50°

To Find:

The measure of ∠CDE.

Solution:

Since ABCD is a parallelogram, its pair of opposite sides are equal, i.e.,

$AB=CD$

$AD=BC$

Given that,

$BC=2AB$

$\frac{AB}{BC}=\frac{1}{2}$

Let $BC=2k$

$AB=k$

then, $CD=k$ and $AD=2k$

Now, $BC=2k$

$BC=BE+CE$

But $BE=CE$ (given)

∴ $BC=2BE$

$2k=2BE$

∴ $BE=k=CE$

In a parallelogram, opposite sides are equal and parallel. AD || BC and $AD=BC$. Let DE be the transversal between AD and BC.

∠ADE = 50°

then ∠DEC = 50° (alternate interior angles)

In ΔECD,

CE = k

CD = k

Hence, it is an isosceles triangle. An isosceles triangle is a triangle in which two sides and the angles opposite to these sides are equal.  

In isosceles ΔECD, CE = CD = k. The angle opposite to side CE is ∠CDE and the angle opposite to side CD is ∠DEC which was calculated as 50°. According to a property of an isosceles triangle, the angles opposite the equal sides are also equal.

If ∠DEC = 50°, then ∠CDE = 50° (equal sides, opposite angles)

Hence, ∠CDE = 50°.

In a parallelogram ABCD, the measure of ∠CDE = 50° where E is the midpoint of side BC.