Given: ∠bcd ≅ ∠edc and ∠bdc ≅ ∠ecd prove: δ bcd ≅ δ edc

 The proof is done below using ASA congruence postulate.

Step-by-step explanation:  We are given that in the figure shown :

∠BCD ≅ ∠EDC and ∠BDC ≅ ∠ECD.

We are to prove that :

Δ BCD ≅ Δ EDC.

From the figure, we can see that the side CD is the common side of both the triangles, ΔBCD and ΔECD.

In triangle BCD and ECD, we have

∠BCD ≅ ∠EDC,

∠BDC ≅ ∠ECD

and CD is the common side.

That is, two angles and the side between them of ΔBCD are congruent to the corresponding angles and the side between them of ΔECD.

Therefore, by ASA (Angle-side-Angle) postulate, we arrive at

Δ BCD ≅ Δ EDC.

Hence proved.

im not quite sure if this is right but i got 5.6 oisutuve

step-by-step explanation: