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Thread: Geometry

  1. #1
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    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 $\displaystyle bx^2 + cx $+d and $\displaystyle b1x^2 $ + c1x + d ,then prove that $\displaystyle a(b-b1)=c1 - $c
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  2. #2
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    Hello, muks!

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

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

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

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

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

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  3. #3
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    Thank you! But about the sum in algebra ?
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  4. #4
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    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 $\displaystyle bx^2 + cx $+d and $\displaystyle b1x^2 $ + c1x + d ,then prove that $\displaystyle a(b-b1)=c1 - $c
    Let $\displaystyle P_1(x) = bx^2 + cx + d$ and $\displaystyle P_2(x) = b_1x^2 + c_1x + d$

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

    By the factor and remainder theorems then, we have

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

    Since they both equal 0, they are equal.

    So $\displaystyle ba^2 + ca + d = b_1a^2 + c_1a + d$

    $\displaystyle ba^2 + ca = b_1a^2 + c_1a$

    $\displaystyle a(ba + c) = a(b_1a + c_1)$

    $\displaystyle ba + c = b_1a + c_1$

    $\displaystyle ba - b_1a = c_1 - c$

    $\displaystyle a(b - b_1) = c_1 - c$.
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