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Thread: Yet another trig excercise

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    Default Yet another trig excercise

    From today's maths exam:


    A: Prove AP = CH

    B: AB = BC. Now prove APCH is a rhombus.

    Very small hint:
    Look for congruent triangles

    Small hint:
    You have to draw a line or two

    Bigger hint:
    What are the properties of cyclic quadrilaterals?

    Hints needed: none
    A took me ~20 mins, including writing the correct proof.
    B took me 2 minutes, because I wasted way too much time in A and I had like every angle = some other angle
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    Are we allowed to use the circle, being a circle without proving it?
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    Quote Originally Posted by Dervish View Post
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    The circle is defined as the circumscribed circle of triangle ABC, you may use that.
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    PC = AH
    Becuase AXH is a right angle.

    Therefore, AP = CH.


    Either that, or BAP = BYC, both are right angles which makes AP = CH?

    ________________________________________
    14:19 < cycrosism> I wonder what she would have done without it
    14:19 < cycrosism> without me*
    Cycrosism is now an it.
    Quote Originally Posted by Dervish View Post
    /Facedesk.

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    Look at below... at the red part.. U will notice that Z/N shape where two lines are parallel and are joined by a diagonal.. therefore the two angles must equal.. if the two angles equal then it is true that the sides are the same lengths and the widths are the same lengths.

    To prove its a rhombus.. well we established fact one above.. therefore using the angles.. we know that the blue angles are the same but are smaller (acute) than the green angles which are the same.. while each side is equal to their opposite. Bam problem solved.

    Attachment 10998
    I am Ggzz..
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    Quote Originally Posted by HarryJames View Post
    PC = AH
    Becuase AXH is a right angle.

    Therefore, AP = CH.


    Either that, or BAP = BYC, both are right angles which makes AP = CH?

    http://dl.dropbox.com/u/10658466/yVpUs.png
    Angles alone do not prove that sides are the same length? (correct me if I'm wrong).

    Quote Originally Posted by ggzz View Post
    Look at below... at the red part.. U will notice that Z/N shape where two lines are parallel and are joined by a diagonal.. therefore the two angles must equal.. if the two angles equal then it is true that the sides are the same lengths and the widths are the same lengths.
    It's not a Z/N shape until you prove it is, which you haven't done yet.
    That you can notice it doesn't mean it has to be that way, there could be a really small difference
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    Quote Originally Posted by Markus View Post
    Angles alone do not prove that sides are the same length? (correct me if I'm wrong).


    It's not a Z/N shape until you prove it is, which you haven't done yet.
    That you can notice it doesn't mean it has to be that way, there could be a really small difference

    Its already proven.. look at the diagram u will see the two 90 degree angles at the bottom.. therefore the lines are parallel and are joined by the diagonal which makes the angle inside on opposite sides equal.. Therefore u get the Z/N Shape.. nothing to prove there.. Also if u noticed.. the two points at the top are colour coded (sorta blue).. My teacher once told me that means the lines will be the same length as long as they're parallel.. All the other points are just black. Plus the largest triangle there is equilateral and all the diagonals meet at the center.. CH is therefore the vertical of the triangle since it meets directly at the center of the triangle so I know CH and PA are parallel.. But yeah the Z/N shape is already there.. no need to prove it at all.

    If u wanted to get in dept:
    Attachment 10999
    Last edited by Brandon; 04-11-2011 at 04:59 PM.
    I am Ggzz..
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    ^ Got it now, thanks. Valid
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