In △abc, point m is the midpoint of ab , point d∈ ac so that ad: dc=2: 5. if aabc=56 yd2, find abmc, aamd, and acmd.

Since point M is the midpoint of AB, then AM=MB.

Consider the area of the triangles ABC and BMC:

where is the height drawn from the vertex C to the side AB.

So,

Now

Also

Now consider the area of the triangles AMD and CMD. Let be the height drawn from the point M to the side AC.

Therefore,

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3