deductive geometry

• Oct 15th 2009, 07:14 AM
thereddevils
deductive geometry
Quadrilateral ABCD is a rhombus and AB is extended to P such that BP=AB . Prove that triangles DAB and CBP are congruent and diagonal BD is parallel to PC .

angle B=angle D
angle p=angle C [Both triangles ABD and PBC are isosceles triangles]

angle B+angle D = angle P + angle c

angle A=180-angle B-angle D
angle B=180-angle P-angle C

angle A=angle B
Hence , triangle DAB is congruent to triangle ABD
Since angle P = angle ABD [corresponding angle] , BD // PC

The straight line that passes through D , parallel to AC meets PC prodiced at R , Prove that \$\displaystyle \angle CRD = 90^o\$ .

I am not really sure bout this part. I cant see that which triangle is congruent to which in this case

THanks !
• Oct 15th 2009, 07:43 AM
Hello thereddevils

Quote:

Originally Posted by thereddevils
Quadrilateral ABCD is a rhombus and AB is extended to P such that BP=AB . Prove that triangles DAB and CBP are congruent and diagonal BD is parallel to PC .

angle B=angle D Which angle B? Which angle D?
angle p=angle C [Both triangles ABD and PBC are isosceles triangles] Which angle C?

angle B+angle D = angle P + angle c This is very confused!

angle A=180-angle B-angle D
angle B=180-angle P-angle C

angle A=angle B No.
Hence , triangle DAB is congruent to triangle ABD This is the same triangle!
Since angle P = angle ABD [corresponding angle] , BD // PC

The straight line that passes through D , parallel to AC meets PC prodiced at R , Prove that \$\displaystyle \angle CRD = 90^o\$ .

I am not really sure bout this part. I cant see that which triangle is congruent to which in this case

THanks !

As you can see, I've found your proof to be very confused.

Here's a summary of how you can prove two triangles congruent.

So, all you need here is to say why:

• \$\displaystyle AB=BP\$
• \$\displaystyle \angle DAB = \angle CBP\$

and you can then say that the triangles are congruent (SAS). Can you do this now?

Here's something on equal angles formed by parallel lines.

Because the triangles are congruent, \$\displaystyle \angle ABD = \angle BPC\$. Can you see why this is sufficient to prove that \$\displaystyle BD \parallel PC\$?

One of the properties of a rhombus is that its diagonals are perpendicular, which you'll find on this page. Can you see why this is enough to show that \$\displaystyle \angle CRD = 90^o\$?

• Oct 15th 2009, 10:54 PM
thereddevils
Quote:

Hello thereddevils

See my comments in red.As you can see, I've found your proof to be very confused.

Here's a summary of how you can prove two triangles congruent.

So, all you need here is to say why:

• \$\displaystyle AB=BP\$
• \$\displaystyle \angle DAB = \angle CBP\$
and you can then say that the triangles are congruent (SAS). Can you do this now?

Here's something on equal angles formed by parallel lines.

Because the triangles are congruent, \$\displaystyle \angle ABD = \angle BPC\$. Can you see why this is sufficient to prove that \$\displaystyle BD \parallel PC\$?

One of the properties of a rhombus is that its diagonals are perpendicular, which you'll find on this page. Can you see why this is enough to show that \$\displaystyle \angle CRD = 90^o\$?