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Math Help - rhombus

  1. #1
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    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)
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    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?

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

    Grandad
    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 .
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  4. #4
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    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 .
    I'm not quite sure what you mean here. So can I clarify it.

    In general, the diagonals of a parallelogram and of a rectangle will not be perpendicular. It's only when the parallelogram is a rhombus, or the rectangle a square, that the diagonals will be perpendicular.

    Grandad
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