GE bisects DGF , m_ CGD= 2x - 2, m2 EGF= 37, m CGF = 7x + 2 mZCGD = mZCGF =
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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.
