A regular triangle lies at the base of the prism. Angle A1AB = A1AC. Prove that

| September 19, 2021 | education

A regular triangle lies at the base of the prism. Angle A1AB = A1AC. Prove that 1) BC is perpendicular to AA1; 2) BB1C1C rectangle face

Let’s build the height АН of the triangle ABC. Since the triangle ABC is equilateral, the height of AH is also the stiffness of its median and bisector.

The projection of the edge АА1 onto the plane ABC is the segment AK, which lies at the height АН.

Since AH is perpendicular to BC, then AK is perpendicular to BC, and therefore AA1 is perpendicular to BC, one hundred and it was required to prove.

The side faces of АА1В1В and АА1С1С1 are equal parallelograms, since their opposite sides are equal and the angle BAА1 = САС1. Then BB1 = CC1 and parallel to each other, BC = B1C1 and parallel to each other.

AA1 is perpendicular to BC, and then both BB1 and CC1 are perpendicular to BC, and the quadrilateral BCC1B1 is a rectangle, which was required to prove.



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