ABCD is a parallelogram. BT bisects $\angle ABC$ and meets AD at T. A straight line through C and parallel to BT meets AB produced at P and AD produced at R. Prove that $\Delta RAP$ is isosceles and the sum of two equal sides of the $\Delta RAP$ is equal to the perimeter of the parallelogram ABCD.

Hint: We will use the mid-point theorem to show that $DC=\dfrac{1}{2}AP$ and $CB=\dfrac{1}{2}RA$. Then we will prove that ABCD is a square, then finally we will show that $AP=AR$ by comparing $CB=\dfrac{1}{2}RA$ in square ABCD. So we will prove $AP=AR$ which will show that triangle RAP is an isosceles triangle and $AB+BC+CD+AD=AP+AR$.

Complete step-by-step solution -
It is given in the question that ABCD is a parallelogram. BT bisects $\angle ABC$ and meets AD at T. A straight line through C and parallel to BT meets AB produced at P and AD produced at R. And we have been asked to prove that $\Delta RAP$ is isosceles and the sum of two equal sides of the $\Delta RAP$ is equal to the perimeter of the parallelogram ABCD.

We know that according to the mid-point theorem, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
Also, from the figure, we can say that the side $DC\parallel AB$ and side $DC\parallel AP$. So, on the basis of the mid-point theorem, we can say that,
Side DC = $\dfrac{1}{2}$ Side AP.
We can also write it as,
$DC=\dfrac{1}{2}AP.........\left( i \right)$.
From this, we get $DC=AB=BP$ ---(1).
Similarly, $CB\parallel DA$ and $CB\parallel RA$, we get,
$CB=\dfrac{1}{2}RA.........\left( ii \right)$.
We also get that $CB=DA=DR$ ---(2).
Now, let us assume that ABCD is a square because every square is a parallelogram. So, we get,
$DC=CB=BA=AD$ ---(3).
So, from here we have $DC=CB$. So, on comparing equations (i) and (ii), we get that,
$\dfrac{1}{2}AP=\dfrac{1}{2}AR$.
On cancelling the like terms on both the sides, we get,
$AP=AR$.
Now, we know that in an isosceles triangle, any two sides are equal. So, from our observations, we can say that in $\Delta RAP$ has two sides, AP and AR equal. Hence, $\Delta RAP$ is an isosceles triangle.
From the figure, we can see that $AP=AB+BP$.
From equation (1), we can see $AB=BP$, and from equation (3), we can see $AB=BC$.
So, we can write $BC=BP$. From this, we get $AP=AB+BC$.
From the figure, we can see that$AR=AD+DR$.
From equation (2), we can see $AD=DR$, and from equation (3), we can see $AD=DC$.
So, we can write $DC=DR$. From this, we get $AR=AD+DC$.
So, perimeter of parallelogram ABCD = $AB+BC+CD+DA$.
We also know that,
$\begin{align}
  & AP=AB+BC.........\left( iii \right) \\
 & AR=AD+DC.........\left( iv \right) \\
\end{align}$.
So, on adding equations (iii) and (iv), we get,
$AP+AR=AB+BC+CD+DA$.
We know that $AB+BC+CD+DA$ is the perimeter of the parallelogram ABCD and $AP+AR$ is the sum of two equal sides of an isosceles triangle RAP.
So, we have proved that the sum of two equal sides of the $\Delta RAP$ is equal to the perimeter of the parallelogram ABCD.

Note: The most common mistake that the students make while solving this question is, by interchanging the values of equations (i) and (ii), they may write it as, $\dfrac{1}{2}DC=AP$ and $\dfrac{1}{2}CB=RA$ which is not correct and results in the wrong answer. Also, it is required that the students draw a figure based on the conditions given in the question, only then will they be able to solve the question properly without any errors.