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.

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

ummm seems that u didnt post the attachement !

to be honest i don't know but i need points so