As shown in the figure, in △ ABC, ab = AC, ad is high, and am is the bisector of ∠ CAE at the outer angle of △ ABC. (1) draw the bisector DN of ∠ ADC with ruler; (2) let DN and am intersect at point F to judge the shape of △ ADF. (only write the result)

As shown in the figure, in △ ABC, ab = AC, ad is high, and am is the bisector of ∠ CAE at the outer angle of △ ABC. (1) draw the bisector DN of ∠ ADC with ruler; (2) let DN and am intersect at point F to judge the shape of △ ADF. (only write the result)

(1) As shown in the figure: (2) the shape of △ ADF is isosceles right triangle, the reason is: ∵ AB = AC, ad ⊥ BC, ∵ bad = ∠ CAD, ∵ AF bisects ∵ EAC, ∵ EAF = ∵ fac, ∵ fad = ∵ fac + ∵ DAC = 12 ∵ EAC + 12 ∵ BAC = 12 × 180 ° = 90 °, that is, ? ADF is right triangle, ? AB = AC, ∵ B = ∵ ACB, ∵ EAC = 2 ∵ EAF = ∵ B + ∵ ACB, ∵ EAF = ∵ B, ∵ AF ∥ BC D = ∠ FDC, ∵ DF bisection ∠ ADC, ∵ ADF = ∠ FDC = ∠ AFD, ∵ ad = AF, that is right triangle ADF is isosceles right triangle