Taking the high ad on the hypotenuse BC of an isosceles right triangle as a crease, we fold △ ADB and △ ADC into two mutually perpendicular planes, and prove that: (1) BD ⊥ AC; (2) ∠ BAC = 60 ° Using vector method

Taking the high ad on the hypotenuse BC of an isosceles right triangle as a crease, we fold △ ADB and △ ADC into two mutually perpendicular planes, and prove that: (1) BD ⊥ AC; (2) ∠ BAC = 60 ° Using vector method

(1) This problem with the vector method to some trouble, it is recommended to use the comprehensive method
Prove: ad ⊥ BD, ad ⊥ CD, so ∠ BDC is the plane angle of dihedral angle b-ad-c, so ∠ BDC = 90 degrees
So BD ⊥ CD. And BD ⊥ ad
So BD ⊥ face ADC so BD ⊥ AC
(2) Connect BC, in the original △ ABC, ab = AC, ad ⊥ BC, so BD = DC, let BD = a, then BC = √ 2A
AB = AC = √ 2a, so △ ABC is an equilateral triangle, so ∠ BAC = 60 degree

In ACD of right triangle, ∠ C is 90 degrees. CD = 10. ADC is 30 degrees
Ad = CD / cos 30 degrees
Cos 30 degree = √ 3 / 2, what is √ 3 / 2 and how?
CD = 10, how long is ad?

(20√3)/3
If one angle of RT triangle is 30 degree, the opposite side of 30 degree angle is half of the hypotenuse
If the opposite side is 1, the hypotenuse is 2, and the other right angle side is √ (2 ^ 2-1 ^ 2) = √ 3
Cos30 = adjacent / hypotenuse = √ 3 / 2
I've answered you twice. Just remember that