Results 1 to 4 of 4

Math Help - Geometry

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    27

    Geometry

    1. P is a point on side BC of a square ABCD. A perpendicular is drawn at AP from the point B which cuts DC at Q.Prove that AP=BQ.(please explain with diagram).Thank you!

    2. If (x-a) be the HCF of  bx^2  + cx +d and  b1x^2 + c1x + d ,then prove that  a(b-b1)=c1 - c
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,911
    Thanks
    773
    Hello, muks!

    1. P is a point on side BC of a square ABCD.
    From point B, a perpendicular is drawn to AP, cutting DC at Q.

    Prove: . AP=BQ
    Let BQ intersect AP at R.
    Code:
                    Q
        D * - - - - * - - - - * C
          |          *        |
          |           *       |
          |            *      |
          |             *     |
          |              *    *P
          |              R*  β|
          |           *    *  |
          |       *         *α|
          |   * α          β *|
        A * - - - - - - - - - * B

    In right triangle ARB, let \angle RAB = \alpha
    Then \angle RBA = \beta, the complement of \alpha
    Hence, \angle RBP = \alpha, the complement of \beta

    Right triangles PBA\text{ and }QCB have: . \begin{Bmatrix}\text{an equal side:} & AB \:=\:BC \\ \text{an equal angle: } & \angle PAB \:=\:\angle QBC \end{Bmatrix}

    Therefore: . \Delta PBA \cong \Delta QCB \quad\Rightarrow\quad AP \:=\:BQ

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    27
    Thank you! But about the sum in algebra ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    Quote Originally Posted by muks View Post
    1. P is a point on side BC of a square ABCD. A perpendicular is drawn at AP from the point B which cuts DC at Q.Prove that AP=BQ.(please explain with diagram).Thank you!

    2. If (x-a) be the HCF of  bx^2 + cx +d and  b1x^2 + c1x + d ,then prove that  a(b-b1)=c1 - c
    Let P_1(x) = bx^2 + cx + d and P_2(x) = b_1x^2 + c_1x + d

    You have said that x-a is the highest common factor between them. So it is a factor of them both.

    By the factor and remainder theorems then, we have

    P_1(a) = ba^2 + ca + d = 0 and P_2(a) = b_1 a^2 + c_1 a + d = 0.

    Since they both equal 0, they are equal.

    So ba^2 + ca + d = b_1a^2 + c_1a + d

    ba^2 + ca = b_1a^2 + c_1a

    a(ba + c) = a(b_1a + c_1)

    ba + c = b_1a + c_1

    ba - b_1a = c_1 - c

    a(b - b_1) = c_1 - c.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Finite Geometry: Young's Geometry
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: September 15th 2010, 08:20 AM
  2. Geometry (1)
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 7th 2009, 07:44 PM
  3. geometry (2)
    Posted in the Geometry Forum
    Replies: 4
    Last Post: November 7th 2009, 07:41 PM
  4. geometry (3)
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 7th 2009, 08:22 AM
  5. Modern Geometry: Taxicab Geometry Problems
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 30th 2009, 08:32 PM

Search Tags


/mathhelpforum @mathhelpforum