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An isosceles triangle ABC, in which AB = BC = 6√2 and AC = 12 is folded along the altitude BD
so that planes ABD and BDC form a right dihedral angle. Find the angle between side AB and its
new position.
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The solution by the other tutor (which, I believe, is an AI solution) is FATALLY WRONG.
I came to bring a correct solution.
In the isosceles triangle ABC
AB^2 + BC^2 = + = 36*2 + 36*2 = 72 + 72 = 144
and
AC^2 = = 144.
It means that triangle ABC is a right-angled isosceles triangle with the right angle at B,
hypotenuse BC and the legs AB and BC.
Now imagine that the triangle ABC is folded along the altitude BD so the planes ABD and BDC form a right dihedral angle.
Let plane BDC remains in its original position and vertex A takes new position A' in space.
Then the segments A'D and CD form the right angle.
Consider triangle A'DC. It is a right-angled triangle with the right angle D and the legs A'D and CD.
These legs have equal length A'D = CD = 12/2 = 6, A'D= AD and AD and CD are equal halves of the base AC of the triangle ABC.
Therefore, the hypotenuse A'C of the triangle ADC has the length = = .
Thus the triangle A'BC (after folding) has congruent sides A'B, A'C and BC: the length of each of these sides is .
Thus the triangle A'BC is an equilateral triangle and all its angles have the measure of 60°.
In particular, the angle A'BC is 60°.
The angle between AB and its new position A'B is the same 60°. The proof is the same.
Solved.