Question

This is a proof question. Need answers.

Answer 1

In triangle BCF and DCF, we have ∠B ≅ ∠D (given), ∠C ≅ ∠D (from step 3), and BF ≅ DF (given that AE ≅ FC). Therefore, by the Angle-Angle-Side (AAS) congruence rule, triangle BCF ≅ triangle DCF.

The given information states that:

Cis the midpoint of AE.

Triangle AEF is congruent to triangle FEC (given AE ≅ FC and ∠A ≅ ∠E).

∠B ≅ ∠D.

We need to prove that ∠BCF ≅ ∠DCF.

Here are the steps to solve the problem:

Step 1: Apply the given information:

Since C is the midpoint of AE, we know that AC = CE.

Step 2: Use triangle congruence:

We are given that triangle AEF ≅ triangle FEC. This means that all corresponding angles and sides are congruent.

Therefore, we can conclude that ∠B ≅ ∠C and ∠F ≅ ∠A.

Step 3: Combine the information:

From step 1, we know that ∠B ≅ ∠D and from step 2, we know that ∠B ≅ ∠C. Therefore, we can conclude that ∠C ≅ ∠D.

Step 4: Conclusion:

In triangle BCF and DCF, we have ∠B ≅ ∠D (given), ∠C ≅ ∠D (from step 3), and BF ≅ DF (given that AE ≅ FC).

Therefore, by the Angle-Angle-Side (AAS) congruence rule, triangle BCF ≅ triangle DCF.

Therefore, we have proven that ∠BCF ≅ ∠DCF.

Answer 2

The prove, ∠ BCF ≅ ∠ DCF. is achieved using SAS Transitive property

How to complete the proof

The two column proof is written as follows

Statement Reason

C is the midpoint of AE given

AE ⊥ FC given

∠ A ≅ ∠ E given

∠ B ≅ ∠ D given

AC ≅ CE Definition of midpoint

Δ BAC ≅ Δ DEC AAS congruence theorem

∠ BCA ≅ ∠ DCE CPCTC

∠ ACF ≅ ∠ ECF right angles are equal

∠ BCA + ∠ BCF = ∠ ACF Angle addition postulate

∠ BCF = ∠ ACF -- ∠ BCA Simplifying

∠ BCF = ∠ ECF -- ∠ DCE Substitution property of equality

∠ DCE + ∠ DCF = ∠ ECF Angle addition postulate

∠ DCF = ∠ ECF - ∠ DCE Simplifying

∠ BCF ≅ ∠ DCF Transitive property