In a parallelogram ABCD, the altitude from the vertex B to the side AD bisects side AD. The perimeter of ABCD is 38 in, and the perimeter of △ABD is 10 in less. Find BD, AB, and AD.

Answer:DB = 9 in , AB = 9 in , AD = 10 inStep-by-step explanation:* Lets explain how to solve the problem- The figure ABCD is a parallelogram - The altitude from B to the side  AD  bisects it- In Δ DAB∵ BE ⊥ AD where E is the point of intersection between the   altitude from point B to the side AD∵ The altitude from point B to the side AD bisects it∴ Point E is the mid-point of AD- In any triangle if a segment drawn from a vertex perpendicular  and bisects the opposite side, then the triangle is isosceles triangle∵ BE ⊥ AD and EA = ED∴ Δ BAD is an isosceles triangle∴ AB = DB- The perimeter of the parallelogram is 38 inches∴ 2 AD + 2 AB = 38 ⇒ divide both sides by 2∴ AD + AB = 19- The perimeter of Δ BAD is 10 less∴ The perimeter of Δ BAD = 38 - 10 = 28∴ AB + DB + AD = 28∵ AB + AD = 19∴ 19 + DB = 28 ⇒ subtract 19 from both sides∴ DB = 9 ∵ AB = DB∴ AB = 9 ∵ AB + AD = 19∴ 9 + AD = 19 ⇒ subtract 9 from both sides∴ AD = 10 ∴ DB = 9 in , AB = 9 in , AD = 10 in