In square ABCD, M is the midpoint of line AB. A line perpendicular to line MC of M meets line AD at K. Prove that angle BCM is congruent to angle KCM?

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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**

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