MHB Problem on Parallelogram proof

[mathlearn]

Can anyone solve the following question

View attachment 5773View attachment 5773

Please try to make you answer detail as possible

:)

 

[MarkFL]

Hello, and welcome to MHB, mathlearn! (Wave)

Our goal here is not to simply provide detailed solutions to posted problems, but rather to help the student by engaging them in the process of arriving to a solution. The student benefits more by doing than by watching.

We expect for you to post what you've tried so far, in order that we can see where you are stuck and how you might be going astray so we can address those points to help guide you in the right direction.

You've posted 3 questions so far without any work shown, so please revisit your 3 threads and include your attempts at solving them and our helpers will be able to provide help aimed specifically at your needs. :)

 

[mathlearn]

S far i have drawn the Line QS other than that i think it has got to to do some thing with the midpoints of the trapezium

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So far i have drawn the Line QS other than that i think it has got to to do some thing with the midpoints of the trapezium

 

[Greg]

Consider triangle $PQS$. What is true about segments $\overline{AD}$ and $\overline{QS}$ ?

 

[mathlearn]

Dear greg1313,

Considering the triangle PQS I think AD should be equal to half of QS. And I am still no seeing a way to prove that ABCD is a parallelogram

 

[mrtwhs]

Am I missing something? An equiangular triangle means that all three angles are equal (in other words, the same as equilateral). In the given problem that is not (in general) going to be true.

 

[Evgeny.Makarov]

Considering the triangle PQS I think AD should be equal to half of QS. And I am still no seeing a way to prove that ABCD is a parallelogram

Yes, $AD=QS/2$, but also $AD$ is parallel to $QS$ as a midsegment in $\triangle PQS$ (the triangle midsegment theorem). Similarly, $BC$ is parallel to $QS$. Therefore, $AD$ is parallel to $BC$. In a similar way one proves that $AB$ is parallel to $DC$. And the fact that opposite sides are parallel is one of the equivalent definitions of a parallelogram.

Actually, the fact that $AD=BC=QS/2$ and $AB=DC=PR/2$ is yet another one of the equivalent definitions of a parallelogram.

Am I missing something? An equiangular triangle means that all three angles are equal (in other words, the same as equilateral). In the given problem that is not (in general) going to be true.

Apparently, the problem asks to show that the two triangles $PST$ and $QRT$ have their respective angles equal, not that all angles are equal in one triangle. In this case, $\triangle PST$ and $\triangle QRT$ are similar, which implies the equality in item v.

 

[mathlearn]

Thank you for your detailed explanation.Going ahead

View attachment 5781

Help me to say $$\triangle PST and\triangle QRT $$are equiangular triangles

and show PS/QR=PT/QT

Many thanks

 

[Evgeny.Makarov]

$\angle PST=\angle QRT$ because these are corresponding angles at parallel lines $PS$ and $QR$. Similarly $\angle SPT=\angle RQT$. Angle $T$ is common to both triangles. Thus, the triangles are similar. This link gives an equivalent definition of similarity in terms of side ratios.