Question

O is any point inside the rectangle ABCD then prove that OB^2 + OD^2 = OA^2 + OC^2

Answer 1

${\mathfrak{\red{\underline{\underline{Solution:-}}}}}$

$\sf{From\;the\;point\;O\;construct\;perpendiculars\;OP,\;OQ,\;OR\;and\;OS}$$\sf{to\;the\;side\;of\;the\;rectangle}$

$\sf{LHS=OA^{2}+OC^{2}}$

$\sf{=(AS^{2}+OS^{2})+(OQ^{2}+QC^{2})\;\;\;(Considering\;right\;triangles\;ASO\;andCOQ)}$

$\sf{But\;AS=BQ\;and\;OQ=BP.\;Therefore,}$

$\sf{(AS^{2}+OS^{2})+(OQ^{2}+QC^{2})}$

$\sf{=(BQ^{2}+OS^{2})+(OQ^{2}+QC^{2})}$

$\sf{=(BQ^{2}+OQ^{2})+(OS^{2}+QC^{2})}$

$\sf{=OB^{2}+OD^{2}\;\;\;\;(Considering\;right\;triangles\;BOQ\;and\;ODS)}$

$\sf{=RHS}$

$\huge{\bf{\underline{Hence\;Proved}}}$