As shown in the figure, take the high ad on the hypotenuse BC of the isosceles right triangle ABC as the crease to make the planes of △ abd and △ ACD perpendicular to each other, The midpoint of BC is n (1) verification: plane ACD vertical plane BDC (2) verification: plane adn vertical plane ABC (3) degree of angle cab

As shown in the figure, take the high ad on the hypotenuse BC of the isosceles right triangle ABC as the crease to make the planes of △ abd and △ ACD perpendicular to each other, The midpoint of BC is n (1) verification: plane ACD vertical plane BDC (2) verification: plane adn vertical plane ABC (3) degree of angle cab

(1) , planar abd ⊥ planar ACD,
BD⊥AD,
⊥ BD ⊥ plane ACD,
∵ BD ∈ plane BDC,
Plane ACD ⊥ plane BDC
(2) , ∵ n is the midpoint of BC,
AB=AC,
⊥ an ⊥ BC, (isosceles triangle with three lines in one),
Similarly, BD = CD,
DN⊥BC,
∵DN∩AN=N,
⊥ BC ⊥ plane adn,
∵ BC ∈ plane ABC,
Plane adn ⊥ plane ABC
(3) Let AB = AC = 1,
Then BD = CD = √ 2 / 2,
And BD ⊥ planar ADC,
CD ∈ planar ADC,
∴BD ⊥CD,
BDC is isosceles RT △,
BC=√2CD=1,
The ABC is positive,
∴

Taking the high ad on the hypotenuse BC of an isosceles right triangle as a crease, the triangle abd and triangle ACD are folded into two mutually perpendicular planes. It is proved that BD is perpendicular to CD and the angle BAC is 60 degrees

1. Angle BDC is the plane angle of dihedral angle, so angle BDC = 90 degree BD vertical CD
2.AD=BD=CD
AB = AC = BC triangle BAC is equilateral triangle, so the angle BAC = 60 degrees

It is known that the quadrilateral ABCD is a right angled trapezoid,
E is the midpoint of PA

(1) Take the midpoint F of PD and connect EF and CF
Then EF / / AD, EF = 1 / 2ad
∵BC//AD,BC=1/2AD
∴BE//=BC
The BCFE is a parallelogram
∴BE//CF
Be is not in PCD and CF is in PCD
Be / / planar PDC
（2）
∵PD=1,PA=3,AD=2√2
∴PD²+AD²=PA²
∴PD⊥AD
∵ planar pad ⊥ planar ABCD, and the intersection line is ad
PD in planar pad
{PD ⊥ plane ABCD
∴AB⊥PD
Abd is an isosceles right triangle
∴AB⊥BD
And BD intersects PD
The PBD of AB vertical plane

Take the high ad on the hypotenuse BC of isosceles triangle ABC as crease, make the angle BDC as right angle, judge whether the plane ADC is vertical to the plane ADB, and explain the reason

Vertical. Because BD is vertical to CD and ad, BD is vertical to ACD. So abd is vertical to ACD (if a line in the plane is vertical to another plane, the plane is vertical to another plane)