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20 15 Xuhui district mathematics module 2
Answer: Proof: (1)∵AD=CD (known),

∴∠DCA=∠DAC (equilateral and equiangular);

∫AC aliquot ∠DAB (known),

∴∠DAC =∞∠cab (the property of angular bisector),

∴∠DCA=∠CAB (equivalent substitution),

∴DC∥AB (internal dislocation angles are equal and two straight lines are parallel);

In Rt△ACB, ∠ ACB = 90, ∠ B = 60.

∴∠ cab = 30 (the two acute angles of a right triangle are complementary),

∴∠DAC=30,

∴∠DAB=30 +30 =60 =∠B,

∴AD=BC (equilateral);

∵∠b+∠DAB = 60+60 = 120≠ 180

∴AD and BC are not parallel, …( 1)

The quadrilateral ABCD is an isosceles trapezoid. ...( 1)

(2) From (1), we know that AD=CD, BC=AD,

∴BC=CD (equivalent substitution);

In Rt△ACB, ∠ ACB = 90, ∠ cab = 30.

∴ BC = 12ab = Be (in a right triangle, the right side facing 30 is half of the hypotenuse);

∴CD=BE (equivalent substitution),

∵DC∨AB (the nature of trapezoid),

∴ Quadrilateral DEBC is a parallelogram (parallelograms with equal opposite sides are parallelograms);

BC = CD,

∴ Quadrilateral DEBC is a diamond (parallelogram with equal adjacent sides is a diamond).