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 $bx^2 + cx$+d and $b1x^2$ + c1x + d ,then prove that $a(b-b1)=c1 -$c

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

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 $bx^2 + cx$+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 $ba^2 + ca + d = b_1a^2 + c_1a + d$

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

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

$ba + c = b_1a + c_1$

$ba - b_1a = c_1 - c$

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