Question

Given: AB ≅ AE; BC ≅ DE Prove: ∠ACD ≅ ∠ADC Complete the paragraph proof. We are given AB ≅ AE and BC ≅ DE. This means ABE is an isosceles triangle. Base angles in an isosceles triangle are congruent based on the isosceles triangle theorem, so ∠ABE ≅ ∠AEB. We can then determine △ABC ≅ △AED by . Because of CPCTC, segment AC is congruent to segment . Triangle ACD is an isosceles triangle based on the definition of isosceles triangle. Therefore, based on the isosceles triangle theorem, ∠ACD ≅ ∠ADC.

Answer 1

1. SAS
2. AD

(Edg) I'm doing the assignment right now and these were the correct answers.

Answer 2

Answer:

In isosceles triangle - the base angles are congruent.

so,

   ∠ACD ≅ ∠ADC

Step-by-step explanation:

Given: AB ≅ AE, BC ≅ DE

If AB = AE

Then,

 ΔABE is a isosceles triangle.

The base angle of isosceles triangle are same.

So, ∠B = ∠E

Now in ΔABC and ΔAED

BC ≅ DE , ∠B = ∠E, AB ≅ AE

From side angle side theorem

ΔABC ≅ ΔAED

Then,

        AC ≅ AD

Hence, ΔACD is a isosceles triangle.

And we know that the base angle of isosceles triangle are same.

So,

    ∠ACD ≅ ∠ADC  proved