prove that the diagonals of a rectangle ABCD are equal and bisect each other
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Answers
Answer:
ANSWER
ΔADC and ΔBDC are right angled triangle with AD and BC are hypotenuse.
AC
2
=AB
2
+DC
2
AC
2
=(5−2)
2
+(6+1)
2
=9+48=58 sq.unit
BD
2
=DC
2
+CB
2
BD
2
=(5−2)
2
+(−1−6)
2
=9+49=58 sq.unit
Hence, both the diagonals are equal in length.
In ΔABO and ΔCDO
Since, ∠OAB=∠OCD, ∠OBA=∠ODC (Both are alternate interior angles of parallel lines)
and AB=CD
Therefore ΔABO≅ΔCDO
⇒AO=CO and BO=DO
Therefore, Both diaginals bisects each other.
Answer:
ΔADC and ΔBDC are right angled triangle with AD and BC are hypotenuse.
AC
2
=AB
2
+DC
2
AC
2
=(5−2)
2
+(6+1)
2
=9+48=58 sq.unit
BD
2
=DC
2
+CB
2
BD
2
=(5−2)
2
+(−1−6)
2
=9+49=58 sq.unit
Hence, both the diagonals are equal in length.
In ΔABO and ΔCDO
Since, ∠OAB=∠OCD, ∠OBA=∠ODC (Both are alternate interior angles of parallel lines)
and AB=CD
Therefore ΔABO≅ΔCDO
⇒AO=CO and BO=DO
Therefore, Both diaginals bisects each other.