P & Q are midpoint of side CA & CB respectively of ΔABC right angled at C. Prove that 4( AQ² + BP² ) = 5AB²
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Answer:
that's all about my answer
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)
