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As shown in the figure, in square ABCD, e is on CD, and AE = EC + BC, M is the midpoint of CD, and the proof is: ∠ BAE = 2 ∠ dam The map address is D: / / my documents / doc1.mht
Make the bisector AF of ∠ BAE intersect BC with F, and make FG ⊥ AE with G,
Then FG = FB, Ag = ab
AE = EC + BC, ab = BC
So, eg = EC
Therefore, after connecting Fe, there is △ GEF ≌ △ CFE
So FC = FG
It has been proved that FG = FB
So, FB = FC
F is the midpoint of BC
Therefore, ∠ BAF = ∠ dam
Because AF is the bisector of BAE, BAE = 2 dam
In square ABCD, e is on CD, and AE = EC + BC, M is the midpoint of CD. Prove the angle BAE = 2 angle dam
…… It's better to write the process clearly. I'm sorry you don't understand that idea = =
It is proved that extending DC to h makes ch = BC, connecting ah to BC at o point. ∵ ch = AB AE = EC + BC quadrilateral ABCD is a square ∵ AE = eh, i.e. ∵ eah = ∵ h ∵ ab ∥ DH ∵ H = ∵ HAB ∵ HCO = ≌ B = 90 ° Ch = ab ≌ HCO ≌ ABO, i.e. ≌ Bo = CO and ∵ m is the midpoint of DC ∵ Bo = DM
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