Question

in the given figure,ABCD is a parallelogram angle ADE=50 degree and Angle ACE angle BED is 90 degree.The value of angle EAC+ angle ABC-2 angle DAC is

Answer 1

Given: AB C D is a parallelogram, such that ∠AD E =50°, ∠ AC E= ∠ B ED=90°

To find: ∠ EA C + ∠ A BC - 2 ∠DA C

Solution: In parallelogram AB CD

        ∠ AC E= ∠ B ED=90°

∴ ∠ CA B = ∠ E BA = 90° [ opposite angles of a parallelogram are equal]

∵Parallelogram A B CD is a rectangle.

In Δ ACE and ΔBC E

AC =BE [opposite sides of rectangle]

∠ ACE = ∠B EC  [each being 90°]

A E = B C [ Diagonals of a rectangle are equal]

Δ ACE ≅ ΔBC E  [SSS]

∠ C A E  = ∠ E B C  [C P CT]

In ΔDA C

∠ D + ∠ DA C + ∠ AC D =180°[ angle sum property of triangle]

50° +∠ DA C  + 90° =180°

∠ DA C =180° - 90°-50°

∠ DA C =40°

now, ∠ EA C + ∠ A BC - 2 ∠DA C

= ∠ E B C + ∠ A BC - 2 ∠DA C  [ As,∠ C A E  = ∠ E B C]

= ∠A BE - 2× 40°

= 90° - 80°

=10°

So,the value of ,∠ EA C + ∠ A BC - 2 ∠DA C=10°

Answer 2

Thank you for asking this question. Here is your answer:

We know that the Angle BED = 90°

And Angle AEC = 45°

Then the Angle EAB = 45°

Angle ABC is equal to 45°

Angle ADE = 50°

And Angle DAC = 40°

Angle EAC+ Angle ABC - 2 Angle DAC =  45° +45°-2×40°

= 90° - 80°

= 10°

So the final answer for this question is 10°

If there is any confusion please leave a comment below.