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Point P  is inside rectangle ABCD . Show that

 

PA^2 +PC^2 =PB^2 + PD^2.

Be sure that your proof works for ANY point inside the rectangle.

 Jul 11, 2018

Best Answer 

 #1
avatar+21848 
+1

Point P  is inside rectangle ABCD . Show that

PA^2 +PC^2 =PB^2 + PD^2.

 

 

\(\begin{array}{|rcll|} \hline PA^2 &=& x_1^2+y_1^2 & PC^2 = y_2^2+x_2^2 \\\\ PB^2 &=& y_2^2+x_1^2 & PD^2 = x_2^2+y_1^2 \\\\ \hline PA^2 + PC^2 &=& x_1^2 + y_1^2 +y_2^2+x_2^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\\\ PB^2 + PD^2 &=& y_2^2 + x_1^2 +x_2^2+y_1^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\ \hline \end{array}\)

 

laugh

 Jul 12, 2018
 #1
avatar+21848 
+1
Best Answer

Point P  is inside rectangle ABCD . Show that

PA^2 +PC^2 =PB^2 + PD^2.

 

 

\(\begin{array}{|rcll|} \hline PA^2 &=& x_1^2+y_1^2 & PC^2 = y_2^2+x_2^2 \\\\ PB^2 &=& y_2^2+x_1^2 & PD^2 = x_2^2+y_1^2 \\\\ \hline PA^2 + PC^2 &=& x_1^2 + y_1^2 +y_2^2+x_2^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\\\ PB^2 + PD^2 &=& y_2^2 + x_1^2 +x_2^2+y_1^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\ \hline \end{array}\)

 

laugh

heureka Jul 12, 2018

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