Question

Given:

Answer 1

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.