中四數學科幾何學問題。急急急

Let P be a point on AB such that PD // BC, then BCDP is a parallelogram.
PD = BC (prop. of parallelogram)
By given BC = AD , we have
PD = AD , so △ADP is isos.△
so we have ㄥDAP = ㄥDPA and = ㄥCBA(since PD // BC)
thus ㄥDAP = ㄥCBA
In △ABD and △ABC ,
AD = BC (given)
AB = AB(common side)
ㄥDAP = ㄥCBA(proved)
△ABD =~= △ABC (S.A.S.)
Thus ㄥCAB = ㄥDBA ,
∴ △ABE is an isosceles △.

∵AD=DC=CB(given)
DB=CA (prop. of quadrilateral)
DB and CA touch each other at point E
DE=CE, AE=EB (prop. of quadrilateral)
∴△ABE is an isosceles triangle.