Answer:
option B. MB/AM=NC/AN
Step-by-step explanation:
we know that
The <u><em>Triangle Proportionality Theorem</em></u> states that if a line is parallel to one side of a triangle and it intersects the other two sides, then it divides those sides proportionally
In this problem
MN is parallel to BC
MN intersect AC and divide into AN and NC
MN intersect AB and divide into AM and MB
so
Applying the Triangle Proportionality Theorem
Rewrite
Answer:
The attachment shows ΔBAC ~ ΔBDA
Step-by-step explanation:
You want segment AB to be part of two similar, but not congruent, triangles. One way to do that is to make AB the hypotenuse of one triangle and the leg of another.
It is convenient to construct these triangles using point M as the arbitrary midpoint of the hypotenuse of the larger triangle. (We don't know the coordinates of M—we just know it is on the perpendicular bisector of AB.) BC is a diameter of circle M, and AD is the altitude of ΔABC.
