Question

in the diagram, triangle ABC is similar to triangle EDC. a. is line AB parallel to line DE? b. show that triangle ACD is similar to triangle ECB. c. find the measure of angle CAD. d find line ED. e. find line AD explain your reasoning

Answer 1

The measure of the angle CAD is 40 degrees, the length of ED is 12 units, and the length of AD is 12 units.

What is the similarity law?

It is defined as the law to prove that the two 2-dimensional have the same shape, but it is not compulsory to have the same size. The ratio of the corresponding sides is in the same proportions and the corresponding angles are congruent.

Triangle ABC ~ EDC

Angle ACD = Angle BCE = 90 degree  (vertically opposite angle)

Angle ACB = Angle DCE = 90 degree  (vertically opposite angle)

BE ║ AD

Angle EBC = Angle CDA = 40 degree  (alternative angles)

Triangle ABC ~ Triangle EDC

Angle ABC = Angle CDE

Angle BAC = Angle DEC

∴ AB ║ DE

Triangle ACD ~ Triangle ECB

Angle EBC = Angle CDA and

Angle BEC = Angle CAD

∴ Triangle ACD ~ Triangle ECB

Angle A + Angle E = 180 degree

∠x + 50 + 50 + ∠x = 180

∠x = 40

Angle CAD = 40 degree

We know,

Angle A = Angle E = Angle B = Angle D = 90 degree

So ABCD is a square

AB = ED = 12

AD = AB  = BE = ED = 12

Thus, the measure of the angle CAD is 40 degrees, the length of ED is 12 units, and the Length of AD is 12 units.

Learn more about the similarity here:

brainly.com/question/8045819

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