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Quadrilateral ABCD satisfies both conditions below:

 

the translation that maps A to D also maps B to C.

the reflection over line BD maps A to C.

 

Bob claims quadrilateral ABCD must be a square, Prove or disprove his claim.

 

My work so far: I know a translation is a rigid transformation, so if we had a translation that mapped one point to another, i think it means that AB and DC are parallel. For the second condition, it could still be a rhombus! So, I think that his claim is wrong and that a rhombus is another possible configuration. Is this correct? Please tell me if I have any mistakes.

 Jun 2, 2021
 #1
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As clarification, I would say a rhombus satisfies the conditions since they have equal parallel sides (so when one vertex is mapped to its corresponding one, the other will also. For the second condition, rhombus have reflection all symmetry along their diagonals, directly implying the result.

 

same guest here btw :)

 Jun 2, 2021
 #2
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Hi Guest, your logic sounds good to me.   laugh

 

It can definitely be a square or a rhombus.  

 

I think the translation condition means that opposite sides must be parallel and equal length.

The second condition means that the diagonals must intersect at right angles.

 

there are no other quadrilaterals that meet both these conditions.

 Jun 2, 2021
 #3
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Oh, I only just saw the 'same guest here' bit.

It is really good that you are discussing your own problem.  We really like that.

Why don't you become a member and get known to us.  laugh

 

Members who make a good impression get preferential treatment. wink

 Jun 2, 2021

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