Given: AB ≅ BC and MA ≅ PC ∠AMO ≅ ∠CPO Prove: ∆AMO ≅ ∆CPO

OM = ON by using concurrency of Δs BOM and DON

Step-by-step explanation:

In the figure ABCD:

∵ AB // CD

∵ CB // AD

∵ In any quadrilateral If every two sides are parallel then

 it will be a parallelogram

∴ ABCD is a parallelogram

∴ AC and BD bisects each other at O ⇒ (properties of parallelogram)

∴ OD = OB ⇒ (1)

∵ BC // AD

∴ m∠CBD = m∠ADB ⇒ alternate angles

∵ M ∈ BC , N ∈ AD , O ∈ BD

∴ m∠MBO = m∠NDO ⇒ (2)

∵ BD intersects MN at O

∴ m∠MOB = m∠NOD ⇒ (3) (vertically opposite angles)

From (1) , (2) and (3)

∴ ΔBOM ≅ ΔDON

∴ OM = ON

b is the final answer! : )

898.1666667

step-by-step explanation:

5389/6 = 898.1666667