(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.