Suppose that abcd is a trapezoid, with ab parallel to cd, and diagonals ac and bd intersecting at p. explain why (a) triangles abc and abd have the same area; (b) triangles bcp and dap have the same area; (c) triangles abp and cdp are similar; (d) triangles bcp and dap need not be similar.

A trapezoid ABCD is show below,

AB || CD

AC and BD are diagonals intersecting at P.

(a)

In Δ ABC and ABD,

Area of triangle ABC = Area of triangle ABD

Because,

Area of triangle ABC = 1/2 × Base × Height

Base = AB

Height = h, which is same for both the triangle that's why,

Area of triangle ABC = Area of triangle ABD

(b)

As we know that,

Area of triangle ABC = Area of triangle ABD

On subtracting the area of triangle ABP from both the triangles, we get

Area of triangle ABC - area of triangle ABP = Area of triangle ABD - area of triangle ABP

⇒ area of triangle BCP = area of triangle DAP

Hence, triangles BCP and DAP have the same area.

(c)

In triangle ABP and CDP,

∠ABP = ∠CDP (opposite interior angles are equal between two parallel lines)

∠APB = ∠CPD (opposite angles are equal)

∠BAP = ∠DCP (opposite interior angles are equal between two parallel lines)

Therefore,

By AAA similarity

Triangles ABP and CDP are similar.

(d)

In triangle BCP and DAP,

Only ∠BCP = ∠APD (opposite angles are equal).

Therefore,

Triangles BCP and DAP need not be similar.

based on context clues you gave i'm assuming the answer would be c.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x