As shown in the figure, in △ ABC, ∠ a = 50 ° AB = AC, and the vertical bisector De of AB intersects AC at D, then the degree of ∠ DBC is () A. 15°B. 20°C. 30°D. 25°

As shown in the figure, in △ ABC, ∠ a = 50 ° AB = AC, and the vertical bisector De of AB intersects AC at D, then the degree of ∠ DBC is () A. 15°B. 20°C. 30°D. 25°

The solution is known, ab = AC ﹥ a = 50 °, ab = AC ﹥ ABC = ﹥ ACB = 65 ° and ∵ De is vertical and bisecting ab ﹥ DB = ad ﹥ abd = ﹥ a = 50 ﹥ DBC = ﹥ ABC - ﹥ abd = 65 ° - 50 ° = 15 °. So a is selected

As shown in the figure, ab = ad, AC = AE, ∠ DAB = ∠ CAE, be and DC intersect at point P

It is proved that: am ⊥ DP, m ⊥ PE, n ⊥ DAB, CAE (known), and ⊥ DAB + ⊥ BAC = ⊥ CAE + ⊥ BAC (properties of the equation), that is, ⊥ DAC = ⊥ BAE. In △ ADC and △ Abe, ⊥ AB = ad, ⊥ DAC = ⊥ BAE. AC = AE, ⊥ ADC ≌ Abe (SAS) ⊥ DC = be & nbsp; In R T △ amp and R T △ ANP, AP = AP (common edge), am = an (proved), r t △ amp ≌ r t △ ANP (HL), APM = APN (corresponding angles of congruent triangles are equal), PA bisection and DPE (definition of angle bisection line)

As shown in the figure, make isosceles triangle △ abd and isosceles triangle △ ace outside △ ABC, and make their vertex angles ∠ DAB = ∠ EAC, be and CD intersect at point P, and the extension line of AP intersects at point F. try to judge the relationship between ∠ BPF and ∠ CPF and prove it

Reasons: Ag ⊥ CD in G, ah ⊥ be in H, ∵ DAB = ∠ EAC, ∵ DAB + ∠ BAC = ∠ EAC + ∠ BAC, ∵ DAC = ∠ BAE. In △ ADC and △ Abe, ad = ab ⊥ DAC = ∠ baeac = AE, ≌ ADC ≌ Abe (SAS), ≌ DC = be. ∵ Ag ⊥ CD, ah ⊥ be, ≌ Ag = ah