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Math Help - deductive geometry

  1. #1
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    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 .

    Proof : AB=AD=BC [Quadrilateral ABCD is a rhombus]

    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 \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 !
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  2. #2
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    Hello thereddevils

    See my comments in red.
    Quote Originally Posted by thereddevils View Post
    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 .

    Proof : AB=AD=BC [Quadrilateral ABCD is a rhombus]

    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 \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:

    • AD=BC
    • AB=BP
    • \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, \angle ABD = \angle BPC. Can you see why this is sufficient to prove that 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 \angle CRD = 90^o?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    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:

    • AD=BC
    • AB=BP
    • \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, \angle ABD = \angle BPC. Can you see why this is sufficient to prove that 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 \angle CRD = 90^o?

    Grandad
    THanks . This is the continuation of the question .
    Given that LP=6 cm , PN=4 cm , and NM=5 cm , use similar triangles to calculate the length of LN .

    I tried to find 2 triangles which are congruent . Triangle LPN and triangle LNM .

    angle LPN = angle LNM
    angle PLN=angle MLN
    LN = LN

    but one is SAA and the other one is ASA .They cant be congruent .

    Lost again.
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