(1), the intersection point f is the vertical line of EC, and EC intersects with G.

At RT △FAC and RT △FGC,

FC bisects the angle ∠ACB, so ∠ACF=∠DCF,

FC is the common terminal,

So RT △FAC≌RT △FGC.

AC=GC，

It is also known that EC=2AC, so EC=2GC, FG vertically bisects the bottom,

△FEC is isosceles△, ∠E=∠FCE.

∠ADF= 180 -∠FDC

∠FDC= 180 -(∠DFC+∠DCF))

De: ∠ADF=∠DFC+∠DCF

∠DFC= 180 -∠EFC

∠EFC= 180 -(∠E+∠ECF))

Germany: DFC=∠E+∠ECF

So: ∠ADF=∠DFC+∠DCF

=∠E+∠ECF+∠DCF=3∠E

∠ADF is three times that of∠ E.

(2) At △OEQ and △OFQ, the basement OE OF the two deltas is equal to of, and the areas of the two deltas are equal.

Therefore, their * * * homovertex Q is equal to the bottom OE and OF in height.

That is, the distance from the Q point on the straight line OH to both sides of ∠EOF is equal, so the straight line OH is ∠EOF.

Angle dividing line

Then the distance from the point P on the angular bisector OH to both sides of ∠EOF is also equal, △APB and △CPD.

Bottom AB=CD, and the distance between point P and bottom AB and CD is equal, so △APB and △CPD.

Equal in area.

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