In a quadrilateral ABCD, if segment AB is congruent to segment BC and angle A is greater than angle C, then..
a. AD is lesser than CD
b. AD=CD
c. AD is greater than CD
d. cannot be determined
e. none of these
back up your answer
2006-06-11 10:00:00 · 5 answers · asked by michchristine 2 in Science & Mathematics ➔ Mathematics
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
2006-06-11 10:19:38 · answer #1 · answered by just♪wondering 7 · 1⤊ 0⤋
In a triangle, the line area becoming a member of the mid-components of any 2 aspects is parallel to the 0.33 area and is one million/2 of it. So if we've a triangle ABC and a line intersects AB and AC at D and E respectively, then DEIIBC and DE=BC/2
2016-12-08 19:34:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
a. AD < CD
Draw the bisector of the angle B.
If angle A = angle C, the point D would lie on the bisector of B
and AD and CD would be equal
But angle A > angle C, so D lies on the same side of the bisector as A, so AD < CD.
2006-06-11 10:47:03 · answer #3 · answered by rt11guru 6 · 0⤊ 0⤋
the only thing that matches your description is a trapezoid
AB = BC = CD
AC is greater than CD
ANS : C.) AD > CD
The trapezoid is the only shape that is a quadrilateral that can have 3 sides equal, and still have opposite sides that is bigger than the other side.
2006-06-11 10:17:37 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
do your own homework. i have my own to do.
2006-06-11 10:06:47 · answer #5 · answered by Anonymous · 0⤊ 0⤋