# rhombus

• Jan 7th 2010, 03:21 AM
thereddevils
rhombus
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)
• Jan 7th 2010, 05:37 AM
Hello thereddevils
Quote:

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
$AD = BC$ (opp sides of a parallelogram)

$AB = BP$ (given)

$\angle DAB = \angle CBP$ (corresponding angles, $AD \| BC$)
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 $BD \| PR$? And you're given that $AC \| DR$, and $\angle DRC = 90^o$. So that will be enough to prove $DROC$ is a rectangle, and therefore $\angle DOC = 90^o$.

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

Can you fill in all the details now?

• Jan 7th 2010, 06:39 AM
thereddevils
Quote:

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

$AB = BP$ (given)

$\angle DAB = \angle CBP$ (corresponding angles, $AD \| BC$)
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 $BD \| PR$? And you're given that $AC \| DR$, and $\angle DRC = 90^o$. So that will be enough to prove $DROC$ is a rectangle, and therefore $\angle DOC = 90^o$.

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

Can you fill in all the details now?

Thanks Grandad , oh , so the diagonals of the parallelogram are NOT perpendicular to each other , so all these while i am wrong , i thought they are the same , since the diagonals of squares and rectangles are perpendicular to each other .
• Jan 7th 2010, 07:18 AM