Because it has been built to have four right angles.
BE ⊥ BC and CD ⊥ BC by construction, so ∠EBC = ∠DCB = 90°.
BE and CD are both perpendicular to BC, so BE ‖ CD.
ED lies along the line l, and l ‖ BC by construction.
A quadrilateral with both pairs of opposite sides parallel and one angle 90° has all four angles 90°. So BCDE is a rectangle, with BC as its base and BE as its height.
Q2.
BXAE is a rectangle (how?), so the height of the rectangle is the same as the height of the triangle.
Answer
X is the foot of the perpendicular from A to BC, so ∠AXB = 90°. Also ∠XBE = 90° (BE ⊥ BC), and EA lies along l which is parallel to BX.
∠BXA = 90°, ∠XBE = 90°, EA ‖ XB, AX ‖ EB
So BXAE has four right angles → it is a rectangle
Opposite sides of a rectangle are equal, so AX = BE
AX is the height of the triangle and BE is the height of the rectangle. They are equal — which is exactly why the rectangle's height can be used in the triangle's area formula.
Q3.
Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with BC as the base?
Answer
Yes, it still holds. For an obtuse triangle the foot of the perpendicular from A falls outside BC, at D on the line CB extended. Then ∆ABC is the difference of two right-angled triangles, each of which does sit inside a rectangle.
Area (∆ABC) = Area (∆ADC) − Area (∆ADB)
= ½ × h × DC − ½ × h × DB
= ½ × h × (DC − DB)
= ½ × h × BC
Why it happens: ∆ADC and ∆ADB share the same height h = AD, so their areas differ only through their bases DC and DB. Removing the smaller from the larger removes the overlap exactly and leaves ∆ABC, whose base is DC − DB = BC. The formula ½ × base × height therefore covers acute, right-angled and obtuse triangles alike.