Question

ABCD is a parallelogram. BT bisects angle ADC and meets AD produced at T.A straight line through C and parallel to BT meets AB produced at P and AD produced at R. prove that triangle RAP is an isosceles and the sum of two equal sides of traingle RAP is equal to the perimeter of the parallelogram ABCD.

Answer 1

BT bisects $\angle ABC\Rightarrow \angle TBA=\angle TBC=\alpha$ (say)
Since PR || BT,
$\angle TBA=\angle CPB=\alpha$ (Corresponding angles)
$\angle BCP=\angle TBC=\alpha$ (Alternate angles)
$\therefore From\Delta CBP,\angle CBP={180}^{\circ }(\angle CPB+\angle BCP)={180}^{\circ }-2\alpha \angle TAB=\angle CBP={180}^{\circ }-2\alpha In\Delta RAP,\angle RAP\left(Same\angle TAB\right)={180}^{\circ }-2\alpha \angle RPA\left(Same\angle CPB\right)=\alpha \therefore \angle ARP={180}^{\circ }-\{{180}^{\circ }-2\alpha +\alpha \}=\alpha \therefore \angle ARP=\angle RPA=\alpha$
$\Rightarrow \Delta RAP$ is isosceles triangle
Similarly $\Delta DRC$ is isoscels
$\because \angle DR=\angle DRP=\alpha$
$\Rightarrow DR=DC$
$\because AB$ is isoseles $\because \angle ABC=\angle AB$
Also usual side of $\Delta ARP$ are AP & AR
AP + AR
AD + BP + AD + DR
AB + BC + AD + DC
Perimeter of ABCD
$\therefore$ Sum of equal sides $\Delta PAP$
Perimeter of ABCD