(1) The measure of angle BCD is 60 degrees.
(2) AE is parallel to BD
![Image](https://s2.postimage.org/2qqali3fo/33o5q0y.gif)
Here the positions of points B,D,A out of BDAE are fixed.Night reader wrote:
st(1) angle ADC=(360-2*angle BCD)/2=120, angle ADE=60 + angle BDA=60 (or angle ADE/2=60) => angle BDE=120 but angle BAE-? Not sufficient
st(2) AE // BD is the property for many quadrilaterals, Not sufficient for rhombus
Combining st(1&2) angles EAD=ADB, thus angle BAE=BDE=120. Satisfies for both properties of rhombus - // opposite sides and = opposite angles.
IOM C
Your logic is good, but the conclusion is wrong.prachich1987 wrote:Here the positions of points B,D,A out of BDAE are fixed.
Now if we draw a line through point A parallel to BD ,it will meet the extended CD at point E.
So once we draw such parallel line ,the position of E is unique & we can do nothing about it.
So it has to be a rhombus.
Please tell me where I am going wrong
[email protected] wrote:I am getting the answer as C. Could somebody please help me in this question...
Statement 1 alone does not prove that AE is parallel to BD nor does it prove that Triangle AED is an equilateral triangle...
I need both the statements to prove that AEDB is also a Rhombus...
Help needed...
We need both statements:[email protected] wrote:I am getting the answer as C. Could somebody please help me in this question...
Statement 1 alone does not prove that AE is parallel to BD nor does it prove that Triangle AED is an equilateral triangle...
I need both the statements to prove that AEDB is also a Rhombus...
Help needed...
Statement 1: BCD = 60 degrees