Aline segment bk is an angle bisector of δabc. a line km intersects side bc such, that bm = mk. prove: km ∥ ab.

See explanation

Step-by-step explanation:

1. BK is an angle B bisector, then

(definition of angle bisector)

2. BM = MK, then

triangle BMK is isosceles triangle with base BK.

3. Angles adjacent to the base of isosceles triangle are congruent, then

Note that angle MBK is the same as angle CBK.

4. By substitution property,

5. By alternate interior angles theorem,

if , then

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation: