As shown in the figure, in the isosceles right triangle ABC, ∠ ACB = 90 °, ad is the middle line on the waist CB, CE ⊥ ad intersects AB with E

Make CF ⊥ AB in F, intersect ad in G, as shown in the figure, ∵ △ ABC is isosceles right triangle, ∵ ACF = ∠ BCF = 45 °, that is, ∵ ACG = 45 °, B = 45 °, CE ⊥ ad, ∵ 1 + ∠ ace = ∠ 2 + ∠ ace = 90 °, and ∵ 1 = ∠ 2, in △ AGC and △ CEB, ≌ 1 = 2Ac = CB, ACG = CBE, ≌ △ AGC ≌

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1. As shown in the figure, in the isosceles right triangle ABC, ∠ ACB = 90 °, ad is the middle line on the waist CB, CE ⊥ ad intersects AB with E
2. As shown in the figure, ad ∥ BC, ∠ a = ∠ C, be and DF divide ∥ ABC and ∥ CDA equally
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