Answer: The coordinates of the midpoint of diagonal AC are [tex](\frac{a+b}{2} , \frac{c}{2} )[/tex]. The coordinates of the midpoint of diagonal BD are [tex](\frac{a+b}{2} , \frac{c}{2} )[/tex] AC and BD intersect at point E with coordinates [tex](\frac{a+b}{2} , \frac{c}{2} )[/tex]. By the definition of midpoint, CE ≅ EA and BE ≅ ED Therefore, diagonals AC and BD bisect each other.
Step-by-step explanation:
Since, here the coordinates of A ≡ (0,0) B ≡ (a, 0) D≡ ( b,c)
And, here ABCD is the parallelogram.
Therefore AB = CD
If y is the y-coordinate of C Then C≡(a+b, y)
⇒ [tex]a^2 + 0 = (a)^2 + (y-c)^2[/tex]
⇒ y =c
Thus, C≡(a+b, c)
The coordinates of the midpoint of diagonal AC are,
[tex](\frac{a+b}{2} , \frac{c}{2} )[/tex]
And, The coordinates of the midpoint of diagonal BD are,
[tex](\frac{a+b}{2} , \frac{c}{2} )[/tex]
Since, by the below diagram AC and BD are intersecting at point E,
Where the coordinates of E are,
[tex](\frac{a+b}{2} , \frac{c}{2} )[/tex]
Thus, E is the mid point of Both the segments AC and BD
Because, BE = ED and AE = EC
Therefore, we can say that both AC and BD bisect each other.
Answer:
the coordinate point C are (a+b,c)
The coordinates of the midpoint of diagonal AC¯¯¯¯¯ are (a+b/2, c/2)
The coordinates of the midpoint of diagonal BD¯¯¯¯¯ are (a+b/2, c/2)
AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E with coordinates (a+b/2, c/2)
By the definition of midpoint, AE¯¯¯¯¯≅CE¯¯¯¯¯and BE¯¯¯¯¯≅DE¯¯¯¯¯
Step-by-step explanation:
