Math: Area of a Parallelogram equals geometric mean of triangles

All the triangles are similar in the diagram above. What is interesting about it is that

A_parallelogram = sqrt( 2A_1 * 2A_2)

or in terms of the dotted / solid parallelograms:

A_parallelogram = sqrt( P_1 * P_2)

Derivation

(1) The ratio of area of A1 to A2 is

A_1/A_2 = ( L_1 / L_2 )^2 ……subscripts correspond to triangle depicted above

(2) The ratio of area A3 to A2 is:

A_3/A_2 = ( L_3 / L_2 )^2

(3) Base of the big overall triangle:

L_3 = L_1 + L_2

(4) Substitute #3 into #2

A_3/A_2 = (( L_1+L_2) / L_2 )^2

(5) Expand right side

A_3/A_2 = ( L_1^2+2L_1*L_2+L_2^2) / L_2^2 = L_1^2/L_2^2+2L_1*L_2/L_2^2 +L_2^2/L_2^2 = L_1^2/L_2^2+2L_1/L_2+1

(6) The area of the overall big triangle also is related as follows

A_3 = A_p + A_1 + A_2

(7) Substituting #6 in #5:

(A_p + A_1 + A_2)/A_2 = A_p/A_2 + A_1/A_2 + A_2/A_2 = A_p/A_2 + A_1/A_2 + 1 = L_1^2/L_2^2+2L_1/L_2+1

(8) referring back to #1 we can cancel the term on both sides of the equation. The one also cancels on both sides leaving:

A_p/A_2 = 2L_1/L_2

(9) Moving terms around:

A_p = 2 A_2 * (L_1/L_2)

(10) #1 can be rearranged to

sqrt A_1 = sqrt A_2 * ( L_1 / L_2 )

(11) Cutting a term out of #9 equal to #10 yields

A_p = 2 sqrt ( A_1 * A_2 )