Hello thereddevils Originally Posted by

**thereddevils** A parallelogram ABCD with its diagonals meeting at the point O is shown in the diagram . AB is extendedto P such that BP=AB . THe line that passes through D and is parallel to AC meets PC produced at point R and angle CRD =90 .

(1) SHow that the triangles ABD and BPC are congruent .

I managed to prove this .

(2) Show that ABCD is a rhombus .

(3) CR:PC

Not sure bout (2) and (3)

For (1), I assume you used the facts that$\displaystyle AD = BC$ (opp sides of a parallelogram)

$\displaystyle AB = BP$ (given)

$\displaystyle \angle DAB = \angle CBP$ (corresponding angles, $\displaystyle AD \| BC$)

So the triangles are congruent, SAS.

For (2), a parallelogram is a rhombus if its diagonals are perpendicular. Well, can you see how to use a pair of equal angles in the congruent triangles to prove $\displaystyle BD \| PR$? And you're given that $\displaystyle AC \| DR$, and $\displaystyle \angle DRC = 90^o$. So that will be enough to prove $\displaystyle DROC$ is a rectangle, and therefore $\displaystyle \angle DOC = 90^o$.

For (3), use the congruent triangles to show that $\displaystyle PC = BD$. Also, $\displaystyle RC=DO = \tfrac12DB$, to show that $\displaystyle CR:PC=1:2$.

Can you fill in all the details now?

Grandad