(1) extends the CD intersection AB to point f,

∠BDC=∠DBF+∠DFB

=∠DBF+∠BAC+∠C

=2(∠DBF/2+∠BAC/2)+∠C

=2( 180? -∠E)+∠C

=360? -2∠E+∠C

That is, ∠D+2∠E-∠C=360?

(2) extend the intersection of BD and AC at point F.

∠BDC=∠C+∠BFC

∠BDC=∠C+∠B+∠BAF？ ①

And ∠E+∠EDC=∠C+∠EAC.

∠E+∠BDC/2=∠C+∠BAF/2

2∠E+∠BDC=2∠C+∠BAF？ ②

②-① Get 2 ∠ e = ∠ c-∠ b.

That is, ∠ c = ∠ b+2 ∠ e.

Foundation compaction

The sum of two sides of the triangle 13-2<x & lt 13+2 is greater than the third side, and the difference between the two sides is smaller than the third side.

1 1 & lt； x & lt 15

∵x is a positive integer,

∴x= 12、 13、 14

So there are three such triangles.