In a quadrilateral ABCD, prove that ${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}$ , where P and Q are middle points of diagonals AC and BD.

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Solution

Let ABCD be the quadrilateral. Taking A as the origin, let the position vectors of B, C and D be $\vec{b}, \vec{c}$ and $\vec{d}$ , respectively. Then,

Position vector of P = $\frac{\vec{c}}{2}$ (Mid-point formula)

Position vector of Q = $\frac{\vec{b} + \vec{d}}{2}$ (Mid-point formula)

Now,

${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {|\vec{AB}|}^{2} + {|\vec{BC}|}^{2} + {|\vec{CD}|}^{2} + {|\vec{DA}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c} - \vec{b}|}^{2} + {|\vec{d} - \vec{c}|}^{2} + {|\vec{d}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2} - 2 \vec{c} . \vec{b} + {|\vec{b}|}^{2} + {|\vec{d}|}^{2} - 2 \vec{d} . \vec{c} + {|\vec{c}|}^{2} + |\vec{d}| = 2 {|\vec{b}|}^{2} + 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{c} . \vec{d} . . . . . (1)$

Also,

${AC}^{2} + {BD}^{2} + 4 {PQ}^{2} = {|\vec{AC}|}^{2} + {|\vec{BD}|}^{2} + 4 {|\vec{PQ}|}^{2} = {|\vec{c}|}^{2} + {|\vec{d} - \vec{b}|}^{2} + 4 {|\frac{\vec{b} + \vec{d}}{2} - \frac{\vec{c}}{2}|}^{2} = {|\vec{c}|}^{2} + {|\vec{d} - \vec{b}|}^{2} + {|\vec{b} + \vec{d}|}^{2} - 2 (\vec{b} + \vec{d}) . \vec{c} + {|\vec{c}|}^{2} = 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} + 2 {|\vec{b}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{d} . \vec{c} = 2 {|\vec{b}|}^{2} + 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{c} . \vec{d} . . . . . (2)$

From (1) and (2), we have

${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}$

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A B C D

A C^{2} + B D^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2}

If $A B C D$ is a rhombus then prove that ${A B}^{2} + {B C}^{2} + {C D}^{2} + {D A}^{2} = {A C}^{2} + {B D}^{2}$ .

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