If angle BCD measures 70° then so does angle DBC (because you have formed an isoceles triangle inside the larger ΔABC and the two legs are equal so the two angles have to be equal. So we have a two 70° angles which leaves 40° for the 3rd angle, which is ∠BDC.
Since ∠BDC and ∠ADB are supplementary (180°) - that leaves 140° for ∠ADB and is our answer
I am pretty sure it is 29 because i did the math and that is what it comes out to
