ABCD is a rhombus. AC and BD are its two diagonals intersecting at O. Prove that AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2?

May 10, 2018

Please see the explanation.

Explanation:

The sides of rhombus are congruent by definition, the diagonals that are perpendicular bisectors divide the figure into 4 congruent right triangles, se we can conclude the followings:

$\Delta O A B \cong \Delta O B C \cong \Delta O C D \cong \Delta O A D$

$A {B}^{2} = O {A}^{2} + O {B}^{2}$
$B {C}^{2} = O {B}^{2} + O {C}^{2}$
$C {D}^{2} = O {C}^{2} + O {D}^{2}$
$D {A}^{2} = O {A}^{2} + O {D}^{2}$
But:
$A C = O A + O C$
$A {C}^{2} = {\left(O A + O C\right)}^{2} = O {A}^{2} + 2 \cdot O A \cdot O C + O {C}^{2}$
$O A = O C$
$A {C}^{2} = 2 \cdot O {A}^{2} + 2 \cdot O {C}^{2}$
Also:
$B D = O B + O D$
$B {D}^{2} = 2 \cdot O {B}^{2} + 2 \cdot O {D}^{2}$
So:
$A {B}^{2} + B {C}^{2} + C {D}^{2} + D {A}^{2} = 2 \cdot O {A}^{2} + 2 \cdot O {B}^{2} + 2 \cdot O {C}^{2} + 2 \cdot O {D}^{2}$
Thus:
$A {B}^{2} + B {C}^{2} + C {D}^{2} + D {A}^{2} = A {C}^{2} + B {D}^{2}$

Hence proved.