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In square ABCD, M is the midpoint of line AB. A line perpendicular to line MC of M...
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Let consider AB=BC=CD=DA=2x
Then MB= 2x/2 = x since M is the mid point of AB.
Now lets take `angle (BCM) = alpha ` and `angle (KCM)= beta`
So considering triangle CMB
`cos(alpha) = (MB)/(CM) = x/sqrt(2x^2+x^2)= 2/sqrt5`
`tan(alpha) = (MB)/(BC) = x/(2x) = 1/2 `
Now `angle CMB = 90-alpha`
since angle CMK =90 ;
`angle KMA = 180-90-(90-alpha) = alpha`
Considering triangle AMK;
`cos(alpha) = (AM)/(KM)`
` KM = (AM)/cos(alpha) = x/(2/sqrt5) = sqrt5*x/2`
considering triangle KCM ;
`tan(beta) = (KM)/(CM) = (sqrt5*x/2)/(sqrt(2x^2+x^2)) = 1/2`
`tan(alpha) = tan(beta)`
`alpha = beta`
angle BCM = angle KCM
Posted by jeew-m on July 31, 2012 at 9:38 AM (Answer #1)
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