Question 622241
I can't see your figure, so I'll guess it's something like
this.  You have a kite ABCD with a diagonal AC:

{{{drawing(3200/11,370,-1.5,2.5,-3.5,2,green(line(0,1.5,0,-3)),
locate(-.05,1.75,A),locate(-1.2,0.1,B),locate(-.05,-3,C),locate(1.1,0.1,D),
line(-1,0,0,1.5), line(0,1.5,1,0), line(1,0,0,-3),line(0,-3,-1,0) )}}}

Given:   AB ≅ AD
       ∠ABC ≅ ∠ADC
Prove: ᐃABC ≅ ᐃADC

I'll just give you an outline. You'll have to write the
2-column proof.

Draw BD:
 
{{{drawing(3200/11,370,-1.5,2.5,-3.5,2,green(line(0,1.5,0,-3)),red(line(-1,0,1,0)),
locate(-.05,1.75,A),locate(-1.2,0.1,B),locate(-.05,-3,C),locate(1.1,0.1,D),
line(-1,0,0,1.5), line(0,1.5,1,0), line(1,0,0,-3),line(0,-3,-1,0) )}}}

Then ᐃABD is isosceles because its legs AB and AD are given congruent. Therefore 
base angles ∠ABD ≅ ∠ADB.  Since we are given 
∠ABC ≅ ∠ADC, then m∠ABD = m∠ADC.  Since ∠ABD ≅ ∠ADB, m∠ABD = m∠ADB.  
Subtracting equals from equals, we get that m∠CBD = m∠CDB and ∠CBD ≅ ∠CDB.
Therefore ᐃBCD is isosceles because its base angles are congruent.
BC ≅ DC because they are legs of isosceles ᐃBCD.  Now we have 
AB ≅ AD, ∠ABC ≅ ∠ADC, BC ≅ DC.   Then ᐃABC ≅ ᐃADC by SAS.

Edwin</pre>