In a rectangle $A B C D$ , prove that:
$A C^{2} + B D^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2}$

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Solution

Since, $△ A B C$ is a right angled triangle , we have
${A C}^{2} = {A B}^{2} + {B C}^{2}$ -- (1)

Similarly, as $△ B C D$ is a right angled triangle , we have
${B D}^{2} = {C D}^{2} + {B C}^{2}$
But as $B C = A D$ , we have
${B D}^{2} = {C D}^{2} + {A D}^{2}$ -- (2)
Adding eqns 1 and 2, we get
${A C}^{2} + {B D}^{2} = {A B}^{2} + {B C}^{2} + {C D}^{2} + {A D}^{2}$

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

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

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

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

(a) $B C^{2} + A D^{2} + 2 B C . A D$

(b) $A B^{2} + C D^{2} + 2 A B . C D$

(d) $B C^{2} + A D^{2} + 2 A B . C D$

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