On sides CB and CD of square ABCD are chosen points M and K so that the perimeter of triangle CMK equals double the side of the square. Find angle MAK.
Better in my opinion, since it's purely geometric and doesn't use sum of angle tangent formula.
As before AB=1, CM=a, CK=b.
WLOG let ABCD be counterclockwise. Rotate an image of segment DK by 90° clockwise about A so that D'=B and K' is the image of K on line CB and ∠KAK'=90. AK=AK'.
MB=1-a, DK=BK'=1-b, so MK'=2-a-b
From the problem we have a+b+KM=2
so KM=2-a-b and KM=KM'
△AMK ≅ △AMK' by SSS.
∠MAK = 90°/2 = 45°.
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Posted by Jer
on 2023-09-02 17:56:46 |
Better solution
