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Given: DE ≅CE and DC \\ AB. Prove ΔACD ≅ ΔBDC.

Given: DE ≅CE and DC \\ AB. Prove ΔACD ≅ ΔBDC.-example-1

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Statement 2:

Given the triangle CDE:

ΔCDE is isosceles, where DE ≅ EC, so the opposite angles are congruent too:

∠CDE ≅ ∠DCE

Statement 3:

We have:

Then, by alternate interior angles, we conclude that:

∠CDE ≅ ∠EBA

Statement 4:

We have a similar situation as that of the statement 3, so reason is the same to conclude that:

∠DCE ≅ ∠EAB

Statement 5:

We have a similar situation as that of statement 2, so the reason is the same to conclude that:

∠EAB ≅ ∠EBA

Statement 6:

From the previous situation, we see that ΔABE is isosceles, with congruent sides:

EA ≅ EB

Statement 7:

We know that:

EA ≅ EB

ED ≅ EC

AC ≅ EA + EC

BD ≅ EB + ED

Using the two first congruencies on the third one:

AC ≅ EB + ED

Comparing this to the fourth congruence:

AC ≅ BD

So the sum of congruent segments gives us congruent segments.

Statement 8:

The reflexive property states that a quantity is equal to itself. In particular, for the side DC:

DC ≅ DC

Statement 9:

We already know that:

∠CDE ≅ ∠DCE (Statement 2)

DC ≅ DC (Statement 8)

AC ≅ BD (Statement 7)

So, using the Side-Angle-Side theorem, we conclude that ΔACD ≅ ΔBDC

Given: DE ≅CE and DC \\ AB. Prove ΔACD ≅ ΔBDC.-example-1
Given: DE ≅CE and DC \\ AB. Prove ΔACD ≅ ΔBDC.-example-2
Given: DE ≅CE and DC \\ AB. Prove ΔACD ≅ ΔBDC.-example-3