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Daniel [21]
3 years ago
9

Line segment Z E is the angle bisector of AngleYEX and the perpendicular bisector of Line segment G F. Line segment G X is the a

ngle bisector of AngleYGZ and the perpendicular bisector of Line segment E F. Line segment F Y is the angle bisector of AngleZFX and the perpendicular bisector of Line segment E G. Point A is the intersection of Line segment E Z, Line segment G X, and Line segment F Y. Triangle G E F has angles with different measures. Point A is at the center. Lines are drawn from the points of the triangle to point A. Lines are drawn from point A to the sides of the triangle to form right angles and line segments A X, A Z, and A Y. Which must be true? Please Hurry I don't have a lot of time to finsh my test!

Mathematics
2 answers:
ladessa [460]3 years ago
5 0

Answer:

C. Point A is the center of the circle that passes that passes through the points E, F and G and is the center of the circle that passes through the points X, Y and Z.

Step-by-step explanation:

The center of insribed circle into the triangle is the point where the angle bisectors of the triangle meet.

The center of circumsribed circle over the triangle is the point where the perpendicular bisectors of the sides meet.

Line segments ZE, FY and GX are both angle bisectors and perpendicular bisectors of the sides, so the point of intersection of line segments ZE, FY and GX is the center of inscribed circle into the triangle and the center of the circumscribed circle over the triangle. Inscribed circle passes through the points X, Y and Z. Circumscribed circle passes through the points E, F and G. So, point A is the center of the circle that passes that passes through the points E, F and G and is the center of the circle that passes through the points X, Y and Z.

Bumek [7]3 years ago
3 0

Answer:

C

Step-by-step explanation:

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\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

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