Example of a line segment is parallel to a plane

Question:

In the figure shown, P-ABCD is a pyramid. ABCD is a square. PB is perpendicular to the plane ABCD. E is midpoint of PA. F is midpoint of BC. Prove that EB is parallel to plane PDF.

Proof:

Find the midpoint in line segment PD and name this midpoint of PD as point G. Connect EG.

In triangle PAD,

because E is the midpoint of PA and

G is the midpoint of PD,

so, EG // AD and EG = (1/2) AD

This is based on the midpoint theorem. The segment joining the midpoints of two sides of a triangle is parallel to the third side and equals to half the third side.

In square ABCD,

because F is the midpoint of BC.

So, BF // AD and BF = (1/2) AD.

Here we use the property of a square, opposite sides are parallel and equal, interior angles are 90 degrees.

Because EG // AD and BF // AD, so EG // BF.

Because EG = (1/2) AD and BF = (1/2) AD, so EG = BF.

Connect GF,

In quadrilateral EGFB,

Because EG // BF and EG = BF, so, EGFB is a parallelogram.

Here we use the theorem used to prove that a quadrilateral is a parallelogram: If a quadrilateral has a pair of sides that are both parallel and congruent, then the quadrilateral is a parallelogram.

Because EGFB is a parallelogram, so EB = GF.

Here we use the parallelogram theorem, opposite sides of a parallelogram are congruent.

Because GF lies in the plane PDF and EB is not in the plane PDF, so EB // plane PDF.

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