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Math problems in the second day of junior high school (online all day)
Detailed answer:

When the trapezoid is an isosceles trapezoid, CE=4.

Because AEB = CDE+C, ABC = Abd+DBC.

Because the quadrilateral is an isosceles trapezoid, BD=DC.

So ∠AEB=∠ABC, ∠ DBC = ∠ C.

So ∠CDE=∠ABD

It can be proved that △ABD is congruent △DEC(SAS).

CE=AD=4。

When the trapezoid is a right-angled trapezoid, CE=6.

Don't erase that isosceles trapezoid, it helps to do the problem! At this time, e is recorded as m, and e in isosceles trapezoid is still e).

Do DM⊥BC

Because the quadrangle ABDE is an isosceles trapezoid

So ME=2 (you can get it by making two vertical lines along a and d)

Because CE=4 and CM=6, it is actually CE=6.

direct proofs

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