Relevance
• Pope
Lv 7
1 year ago

I get a = 7, b = 9, so a + b = 16.

The way I derived that is a bit sloppy, so instead let me just prove that it satisfies all conditions.

Let line segments AC and BD intersect at P, with these five parameters:

PA = 2

PB = 3

PC = 9

PD = 6

cos(∠BPA) = 7/9

These similar triangles follow:

∆APB ~ ∆DPC, with ratio 1 : 3

∆APD ~ ∆BPC, with ratio 2 : 3

Also, as a result of the similar triangles, ABCD is a cyclic quadrilateral.

AB² = PA² + PB² - 2(PA)(PB)cos(∠BPA)

AB² = 2² + 3² - 2(2)(3)cos(7/9)

AB² = 11/3

AB = √(33)/3

By the similar triangles (∆APB ~ ∆DPC, with ratio 1 : 3),

CD = 3AB

BC² = PB² + PC² - 2(PB)(PC)cos(∠CPB)

BC² = 3² + 9² - 2(3)(9)cos(-7/9)

BC² = 132

BC = 2√(33) = 6AB

By the similar triangles (∆APD ~ ∆BPC, with ratio 2 : 3),

DA = 2/3BC = 4AB

AB : BC : CD : DA

= AB : 6AB : 3AB : 4AB

= 1 : 6 : 3 : 4

Construct midpoints E, F, and G as given. These parallel lines result:

EF || BD

GF || AC

∠EFG = ∠DPC ... (opposite angles of a parallelogram)

∠DPC = ∠BPA ... (vertical angles)

cos(∠EFG) = cos(∠BPA) = 7/9