Is quadrilateral ABCD a rhombus?

Answer D.

(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...

Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

Is quadrilateral ABCD a Rhombus?
(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?

A rhombus is a quadrilateral with all its sides equal to each other.

1.
Imagine that the point where the bisectors meet is called X.
Then, BX = DX and AX = CX.

Using the Pythagorean Theorem you could imagine getting the hypotenuse of sides BX and CX.
Since AX = CX, then the hypotenuse of BX and AX is also as long as the previous.
As you could imagine, all the hypotenuse formed are equal, thus forming equal 4 sides. Hence, a rhombus.

SUFFICIENT.

2.

All sides are equal.

SUFFICIENT.

Answer: D

Hi Bunuel

When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus?

And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus?

Is that the correct line of reasoning?
Thanks so much in advance
Cheers
J

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).

Hope it helps.

But Bunuel this would be true even for rectangle:
Line segments AC and BD are perpendicular bisectors of each other.

I need Your though, am confused. Diagonals of rectangles also bisect each other.

For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°.

Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus.

Hope it's clear.

Bunuel,

For (1), can you consider the case of a kite?

Yes. In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus.

This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus.

Actually, the question is fine.

A rhombus is a quadrilateral all of whose sides are of the same length - that's all. You could have a rhombus which also has all angles 90 which makes it a square or a rhombus in the shape of a kite. But nevertheless, if it is a quadrilateral and has all sides equal, it IS A RHOMBUS.

(1) Line segments AC and BD are perpendicular bisectors of each other.

Make 2 lines - a vertical and a horizontal - which are perpendicular bisectors of each other. Make them in any way of any length - just that they should be perpendicular bisectors of each other. When you join the end points, you will get all sides equal. Think of it this way - each side you get will be a hypotenuse of a right triangle. The legs of the right triangle will have the same pair of lengths in all 4 cases. So AB = BC = CD = AD. So ABCD must be a rhombus.

(2) AB = BC = CD = AD

This statement directly tells you that all sides are equal so ABCD must be a rhombus.

Answer (D)

Superset
The answer to the question will be either yes or no.

Translation
In order to find the answer, we need:
1# measure of length of sides
2# angle between the diagonals
3# other properties to justify that the quadrilateral is a triangle

Statement analysis
St 1: diagonals are bisectors happens in parallelogram. Diagonals are perpendicular happens in kite. A figure common to both is a rhombus. ANSWER.

St 2: if all sides are equal. The figure will be a rhombus. ANSWER

Option D

Aren't the diagonals of any parallelogram perpendicular bisectors?

No.

The diagonals of a parallelogram always bisect (cut in half) each other but they are perpendicular to each other only if a parallelogram is a square.

Hi Bunuel,

Thanks for all your help. I have a question. Why is statement 1 a rhombus and not a kite or square? I am guessing it has to do with intersecting at 90 degrees vs. perpendicular bisectors. What is the difference between the two terminologies?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.

A line segment bisector is a line which cuts a line segment into two equal parts.

As for your other doubt, I think it is answered on the previous two pages:
https://gmatclub.com/forum/is-quadrilat ... ml#p999739
https://gmatclub.com/forum/is-quadrilat ... l#p1277833
https://gmatclub.com/forum/is-quadrilat ... l#p1438436
https://gmatclub.com/forum/is-quadrilat ... l#p1497287

Hope it helps.

Just one question if you could help me.
Are not diagonals of rectangle perpendicular bisectors of each other. I know that they are not equal but they are at 90 degrees to each other and divide the diagonals in equal halves.