# 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
\$\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?

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

Hello thereddevilsFor (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?

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