jmc

Question

Let ABCD be a convex quadrilateral, and let M and N be the midpoints of AC and BD, respectively. Then there exists a constant k so that AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 + k MN^2.Find k. FIGURE

Accepted Answer

Let a = A, etc. Then align* AB^2 &= |a - b|^2 &= (a - b) (a - b) &= a a - 2 a b + b b. align*Similarly, align* BC^2 &= b b - 2 b c + c c, CD^2 &= c c - 2 c d + d d, DA^2 &= d d - 2 d a + a a, AC^2 &= a a - 2 a c + c c, BD^2 &= b b - 2 b d + d d, align*so align* &AB^2 + BC^2 + CD^2 + DA^2 - AC^2 - BD^2 &= a a + b b + c c + d d & - 2 a b + 2 a c - 2 a d - 2 b c + 2 b d - 2 c d. align*Finally, align* MN^2 &= | a + c2 - b + d2 |^2 &= 14 |a + c - b - d|^2 &= 14 (a + c - b - d) (a + c - b - d) &= 14 (a a + b b + c c + d d & - 2 a b + 2 a c - 2 a d - 2 b c + 2 b d - 2 c d). align*Therefore, k = 4.