1. Diagram problem

The diagram consists of a rectangle ABCD and a triangle DXY so that X and Y are points on the line segments AB and BC respectively and angle DXY=90.
Show that DA/XB =AX/BY =XD/YX

Thanks guys.

2. Hello, hongvo!

The diagram consists of a rectangle $\displaystyle ABCD$ and a triangle $\displaystyle DXY$
with $\displaystyle X$ and $\displaystyle Y$ on $\displaystyle AB$ and $\displaystyle BC$, resp. and $\displaystyle \angle DXY\,=\,90^o$

Show that: .$\displaystyle \frac{DA}{XB} \:=\: \frac{AX}{BY} \:=\: \frac{XD}{XY}$
Code:
                X
A o - - - - o - - - - - - - - - o B
|      β *    *   α           |
|       *  90°    *           |
|      *              *       |
|     *                   * β |
|    *                        o Y
|   *                         |
|α *                          |
| *                           |
|*                            |
D o - - - - - - - - - - - - - - o C

$\displaystyle \text{In right triangle }XAD\!: \begin{array}{ccccccc}\text{let}& \alpha &=& \angle ADX \\ \text{let} & \beta &=& \angle AXD &=& 90^o-\alpha \end{array}$

$\displaystyle \text{At }X\!:\;\;\angle AXD + 90^o + \angle BXY \:=\:180^o \quad\Rightarrow\quad \beta + 90^o + \angle BXY \:=\:180^o$

And we have: .$\displaystyle \begin{array}{ccccc}\angle BXY &=& 90^o - \beta &=& \alpha \\ \angle BYX &=& 90^o - \alpha &=& \beta \end{array}$

Hence: .$\displaystyle \Delta XAD \:\sim\:\Delta YBX$

Therefore: .$\displaystyle \frac{DA}{XB} \;=\;\frac{AX}{BY} \;=\;\frac{XD}{XY}$

3. diagram problem

Hi Soroban

Thank you very much for your help. All the best n have a good day.

Cheers, hv

4. sophie diagram problem

Can anyone please tell me if they have solved part c of this question which is:
c. Another sophie diagram has AX=15, YC=16 and XB=48. Find the lengths of the sides of triangle DXY

thanks

5. Hello, imcnalty!

Who is "sophie"?

c. Another sophie diagram has: .$\displaystyle AX\,=\,15,\;\;YC\,=\,16,\;\;XB\,=\,48.$

Find the lengths of the sides of $\displaystyle \Delta DXY.$
Code:
          15    X        48
A o - - - - o - - - - - - - - - o B
|        *    *               |
|       *  90°    *           | b
|      *              *       |
a |     *                   * β |
|    *                        o Y
|   *                         |
|α *                          | 16
| *                           |
|*                            |
D o - - - - - - - - - - - - - - o C
63

Draw $\displaystyle DY.\quad DC = 63.$

Let: .$\displaystyle a \,=\,AD,\;\;b\,=\,BY.$

Note that: .$\displaystyle a \:=\:b+165$ .[1]

In right triangle $\displaystyle XAD\!:\;\;DX^2 \:=\:a^2+15^2$ .[2]

In right triangle $\displaystyle XBY\!:\;\;XY^2 \:=\:b^2 + 48^2$ .[3]

In right triangle $\displaystyle YCD\!:\;\;DY^2 \:=\:16^2 + 63^2 \:=\:4225$ .[4]

In right triangle $\displaystyle DXY\!:\;\;DX^2 + XY^2 \:=\:DY^2$

Substitute [2], [3], [4]: .$\displaystyle (a^2+15^2) + (b^2 + 48^2) \:=\:4225 \quad\Rightarrow\quad a^2+b^2 \:=\:1696$

Substitute [1]: .$\displaystyle (b+16)^2 + b^2 \:=\:1696 \quad\Rightarrow\quad 2b^2 + 32b - 1440 \:=\:0$

. . . . . . . . . . . $\displaystyle 2(b-20)(b+36) \:=\:0 \quad\Rightarrow\quad b \:=\:20 \quad\Rightarrow\quad a \:=\:36$

Therefore: . $\displaystyle \begin{Bmatrix}DX &=&\sqrt{36^2+15^2} &=& 39 \\ XY &=& \sqrt{20^2+48^2} &=& 52 \\ DY &=& \sqrt{4225} &=& 65 \end{Bmatrix}$

6. sophie diagram

hi can you help me with this one to??

DX= 429
DY= 845!!