1. ## 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

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

3. Thank you! But about the sum in algebra ?

4. Originally Posted by muks
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 $\displaystyle ba^2 + ca + d = b_1a^2 + c_1a + d$

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

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

$\displaystyle ba + c = b_1a + c_1$

$\displaystyle ba - b_1a = c_1 - c$

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