Show all the work and make sure to include the formula you used.
2 answers:
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Answer:
(the statement does not appear to be true)
Step-by-step explanation:
I don't think the statement is true, but you CAN compute the intercepted arc from the angle.
Note that BFDG is a convex quadrilateral, so its angles sum to 360. Since we know the inscribed circle touches the angle tangentially, angles BFD and BGD are both right angles, with a measure of 90 degrees.
Therefore, adding the angles together, we have:
alpha + 90 + 90 + <FDG = 360
Therefore, <FDG, the inscribed angle, is 180-alpha (ie, supplementary to alpha)
We know that (-3,5) is the location of one of the endpoints.... and we know the midpoint is at (2,-6)... .now.. what's the distance between those two guys?
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ 5}})\quad % (c,d) &({{ 2}}\quad ,&{{ -6}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[2-(-3)]^2+[-6-5]^2}\implies d=\sqrt{(2+3)^2+(-6-5)^2} \\\\\\ d=\sqrt{5^2+(-11)^2}\implies d=\sqrt{25+121}\implies d=\sqrt{146}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-3%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%202%7D%7D%5Cquad%20%2C%26%7B%7B%20-6%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-3%29%5D%5E2%2B%5B-6-5%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B3%29%5E2%2B%28-6-5%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B5%5E2%2B%28-11%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B25%2B121%7D%5Cimplies%20d%3D%5Csqrt%7B146%7D)
so, the distance "d" from the midpoint to that endpoint is that much. And the distance from the midpoint to the other endpoint is the same "d" distance, because the midpoint is half-way in between both endpoints.
so, the length of AB is twice that distance, or
so, the distance "d" from the midpoint to that endpoint is that much. And the distance from the midpoint to the other endpoint is the same "d" distance, because the midpoint is half-way in between both endpoints.
so, the length of AB is twice that distance, or