Question

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

Answer 1

Step-by-step explanation:

A good diagram helps. We know using Pythagorean theorem that MC is (sqroot5)/2. We also know that angle KMA = angle MCB. That's because CMB is complementary to MCB and 180-(90 +CMB) is MCB!

A good diagram helps. We know using Pythagorean theorem that MC is (sqroot5)/2. We also know that angle KMA = angle MCB. That's because CMB is complementary to MCB and 180-(90 +CMB) is MCB!So now we have similar triangles KAM and CMB. Tan MCB is 1/2. Therefore KA is 1/4 and DK is 3/4. KC is 5/4 using Pythagorean theorem. By the same token KM is (sqroot5)/4. Now figure out tangent of angle KCM using above info and it is also 1/2. Therefore both angles in question are identical.

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Answer 2

Answer:

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