Question

please explain the 17th question clearly correctly with diagram please!!!!

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

$AB= \sqrt{(a-0)^{2}+(0-a)^{2}  } =2\sqrt{a} \\BC= \sqrt{(a-0)^{2}+(0+a)^{2}}=2\sqrt{a} \\ AD=\sqrt{(0-a)^{2}+(a-0)^{2}}=2\sqrt{a}\\BD=\sqrt{(a-0)^{2}+(0+a)^{2}}=2\sqrt{a}\\AB=\sqrt{(a+a)^{2}+(0-0)^{2}}=2a\\CD=\sqrt{(0-0)^{2}+(-a-a)^{2}}=2a\\Now  AB=BC=AD=BD(All \:side\: lengths\ are\: equal)\\and \:also AB=CD(length\: of\: diagonals\: are\: equal)\\Therefore  \:ABCD \: is\:  a \:square\\ Hence \: proved....{statement}$