Known: as shown in the figure, in △ ABC, if ∠ A is an acute angle, points D and E are on AB and AC respectively, and ∠ DCB = ∠ EBC = 12 ∠ A. verification: BD = CE

Known: as shown in the figure, in △ ABC, if ∠ A is an acute angle, points D and E are on AB and AC respectively, and ∠ DCB = ∠ EBC = 12 ∠ A. verification: BD = CE

Method 1: as shown in Figure 1, make CG ⊥ be at G point, make the extension line of BF ⊥ CD at f point. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\in △ CEG, ∠ f = ∠ CGE ∠ Ge Fdbbf = CG, △ BDF ≌ CEG (AAS), and △ BDF ≌ CEG (AAS), and △ BD = CE. Method 2: as shown in Figure 2, taking C as the vertex, make {FCB = DBC, and CF intersection be at point F. in △ BDC and △ CFB, in △ BDC and △ CFB, in △ BDC and △ CFB, ≌ CEG (AAS), CEG (AAS) and ≌ CEG (AAS), as shown in Figure 2, taking C as the vertex of C as the figure 2, as the vertex, make {FCB = as the vertex of C, as the vertex, as the vertex of C, as the vertex of C, as the vertex of C, as the vertex of C, as the vertex as the vertex of C, as the vertex, in △ BDC and △ BDC and △ CFB, in ﹥ a + ﹥ Abe, ﹥ ADC = ﹥ FEC, ﹥ ∠FEC=∠CFE，∴CF=CE，∴BD=CE．