Question

Please help with geometry?

Answer 1

Proof:

Refer to the attached figure.

There are a couple of ways you can go at this. One is to show the sums of the marked angles are the same, hence ∠B ≅ ∠D. Instead, we're going to show that ΔABD ≅ ΔCDB, hence ∠A ≅ ∠C.

1. AB║DC and BC║AD . . . . given

2. BD is a transversal to both AB║DC and BC║AD . . . . given

3. ∠CBD ≅ ∠ADB . . . . alternate interior angles where a transversal crosses parallel lines are congruent

4. ∠CDB ≅ ∠ABD . . . . alternate interior angles where a transversal crosses parallel lines are congruent

5. BD ≅ BD . . . . reflexive property of congruence

6. ΔABD ≅ ΔCDB . . . . ASA postulate

7. ∠A ≅ ∠C . . . . CPCTC

∠A and ∠C are opposite angles of parallelogram ABCD, so we have shown what you want to have shown.