Do the diagonals of a quadrilateral ABCD bisect each other

When we combine (1)+(2), after fixing point A,B,D according to the first condition, C must lie on the perpendicular bisector of BD in order to satisfy BC=CD. Hence (C).

bunuel,

did the problem come with a diagram? i'm a bit confused on how you arrived at a kite. two adjacent sides being equal could be a square or a rhombus.

and I don't know the rule, but it appears that a square/rhombus do not have diagonals that cut at 90 degrees.

thanks for your help!

The diagonals of both square and rhombus cut at 90 degrees.

bunuel, thanks, with that said, can you elaborate on how you visualized the problem and ended at a kite? should the solution encompass that ABCD must be = kite/rhombus/squre and those all have 90 degree intersecting diagonals?

The question is whether the diagonals cut at 90 degrees, we know for sure that ABCD is a kite so we can answer YES to the question. ABCD can also be a rhombus or a square but that's not relevant anymore.

thanks again bunuel! i see what you mean now. the two statements AT MOST tell you it is a kite. whereas we'd require more restrictions for it to be a rhombus, square, etc.

Kinds of quadrilaterals: 1) Square 2) rectangle 3) Trapezoid 4) Rhombus and 5) Kite

well quadrilaterals whose adjacent sides are equal can be 1) Square 2) Rhombus and 3) Kite and all three diagonals cut each other at 90 degrees

so option one tells us that it is either a square or a rhombus or a kite , isn't this sufficient to answer yes to the question ..


similarly option b too tells us that this quadrilateral could be a 1) Square 2) Rhombus or a 3) Kite and all three diagonals cut each other at 90 degrees, so shouldn't be sufficient too

are there any quadrilaterals whose adjacent sides could be equal apart from these 3, whose Diagonals do not cut each other at 90.

In short can someone more elaborate on how either of the statements are insufficient

thanks

Sometimes it's better to draw an actual diagram to test theoretical reasoning. The fact that two sides of a quadrilateral are equal DOES NOT mean that its' either a kite, a square, or a rhombus:You can consider two equal sides to be either AB and AD or BC and DC and see that neither of statement is sufficient on its own.

Hope it's clear.

Hi

I meant two Adjacent sides not any two sides . If two adjacent sides are equal then it must be either a square or a rhombus or a kite , must it not ?

Can you paste a figure of a quadrilateral , with two ADJACENT sides equal , but it must not be a square , a rhombus or a kite , this would certainly make either statements insufficient individually , as any other figure apart from these 3 , will not cut at 90 degrees ,

Thanks

Adjacent sides are sides having have a common vertex. In the diagram in my previous post two adjacent sides are equal (the sides crossed with little line segments).

Yes, It would be better if Bisection was completely removed from this question as we are only testing for perpendicularity.
if the question is asking for both perpendicular and bisection of the Diagonals
using 1+2 we can have the following figures
a kite - no ( here only one Diagonal is Bisected not both )
square - yes
rhombus- no

answer E

If Bisect is removed and only we are checking for diagonals perpendicularity then
using 1+ 2 we can have the following figures
-> square - yes
-> rhombus - yes
--> kite -- yes

Answer C

Members please share your views.Thank you.