Answer:
Step-by-step explanation:
I'm not very familiar with 2 column proofs but the following is how we'd prove it in the UK:
AB ≅ AC ( Given)
BE ≅ CE (Given)
Line segment AE is common to Δ's ABE and ACE
so the 2 triangles are congruent. (By SSS).
< BAD ≅ < CAD ( as triangles ABE and ACE are congruent).
AD is common to Δ's BAD and CAD.
AB ≅ AC (given)
So Δ's BAD and CAD are congruent by SAS.
Therefore < ADB ≅ < ADC ( as triangles BAD and CAD are congruent).
m < BDE = 180 - m < ADB and
m < CDE = 180 - m < ADC (both angles are adjacent)
So since < ADB ≅ < ADC,
< BDE ≅ < CDE.
