Question

ponit T is in the interior of rectangle pqrs prove that TQ square + TS square is equal to TP square + TR square as shown in the figure draw AB parallel to SR

Answer 1

$\textbf{Pythagors theorem:}$

$\text{In a right angled triangle, square on the hypotenuse is equal}$

$\text{to sum of the squares on the other two sides.}$

$\textbf{Given:}$

$\text{PQRS is a rectangle}$

$\textbf{To prove:}$

$TQ^2+TS^2=TP^2+TR^2$

$\text{Apply pythagoras theorem in triangles \triangle SAT, \triangle PAT, \triangle QBT, \triangle RBT}$

$TS^2=SA^2+AT^2$......(1)

$TP^2=PA^2+AT^2$......(2)

$TQ^2=BQ^2+BT^2$......(3)

$TR^2=BR^2+BT^2$......(4)

$\text{(1)-(2) gives}$

$TS^2-TP^2=SA^2-PA^2$.....(5)

$\text{(3)-(4) gives}$

$TQ^2-TR^2=BQ^2-BR^2$

$\boxed{\bf\;BQ=PA\;\text{and}\;BR=SA}$

$TQ^2-TR^2=PA^2-SA^2$

$\implies\;TQ^2-TR^2=-(SA^2-PA^2)$

$\text{Using (5), we get}$

$\implies\;TQ^2-TR^2=-(TS^2-TP^2)$

$\implies\;TQ^2-TR^2=-TS^2+TP^2$

$\text{Rearranging terms, we get}$

$\boxed{\bf\;TQ^2+TS^2=TP^2+TR^2}$

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