Answer:
[tex]\text{Solution:}\\\text{Given: }\\\text{i. M is point outside the plane of rectangle ABCD such that}\\\text{}\ \ \text{ MA=MB=MC=MD,}\\\text{ii. Diagonals AC and BD intersect at O.}\\\text{To prove: MO}\perp\text{plane ABCD, i.e. MO}\perp\text{BD.}\\\text{Proof:}[/tex]
[tex]\text{1. In rectangle ABCD,}\\\text{i. AO=OD and BO=OD \text{[Diagonals of rectangle bisect each other.]}}\\[/tex]
[tex]\text{2. In triangles MOB and MOD,}\\\text{i. MB=MD (S) [Given]}\\\text{ii. BO=OD (S) [From statement 1.i.]}\\\text{iii. OM=OM (S) [\text{Common side in both triangles.}]}\\\text{iv. }\triangle \text{MOB}\cong\triangle\text{MOD [By S.S.S. postulate.]}\\[/tex]
[tex]\text{3. }\angle\text{MOB=}\angle\text{MOD [Corresponding angles of congruent triangles are equal.]}[/tex]
[tex]\text{4. In triangle MBD,}\\\text{i. }\angle \text{MOB+}\angle\text{MOD=180}^\text{o} \text{[Sum of angles in straight line is 180}^\text{o}.]\\\text{ii. }\angle \text{MOB+}\angle\text{MOB=180}^\text{o}\ \ \ [\angle\text{MOB}=\angle\text{MOD}]\\\text{or, }\angle\text{MOB = 90}^\text{o}\\\text{i.e. MO}\perp\text{BD}\\\text{iii. MO is perpendicular to plane ABCD. [From statement 4.ii.]}[/tex]
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