Extend DB to e to make BE=BA, extend DC to f to make CF=CA, and connect AE and AF. From AB+BD=AC+CD, we know that DE=DF, and because AD is perpendicular to BC and △AEF is an isosceles triangle, so ∠AEF=∠AFE, and because △ABF and △ACF are isosceles triangles, so 2∝.

Shit, AB? -BD？ =AC？ DC？ get

(AB+BD)(AB-BD)=(AC+DC)(AC-DC)

So choose [2] [3] [4]