Question

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.

Answer 1

R.E.F image

$\therefore DC\parallel AB$ or $DC\parallel AP$

$\therefore DC=\frac{1}{2}AP...$ by mid-point theorem

$DC=AB=BP$

$CB\parallel AD$ or $CB\parallel RA$

$\therefore CB=\frac{1}{2}RA...$ by mid-point theorem

$CB=DA=DR$

Let $ABCD$ be a square ($\because$ square is also a $\parallel$ gram)

$\therefore DC=CB=BA=DA$

Here, $DC=CB$

and, $DC=\frac{1}{2}AP...\left(i\right)$

$CB=\frac{1}{2}AR...\left(ii\right)$

Comparing (i) and (ii)

$\Rightarrow \frac{1}{2}AP=\frac{1}{2}AR$

Multiplying 2 both sides

$\Rightarrow AP=AR$

$\therefore \Delta RAP$ is an isosceles triangle

Now,

To show :- $AP+AR=$ Perimeter of $\parallel$ gram $ABCD$

We know that,

Perimeter of $\parallel$ gram $ABCD=AB+BC+CD+DA$

but $=AB+BC=AP$ and $AD+DC=AR$ (from ab)

$\therefore AP+AR=$ Perimeter of ABCD

Hence proved.