Diagram problem

**hongvo** · April 28th 2010, 01:28 AM

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.

**Soroban** · April 28th 2010, 06:00 AM

Hello, hongvo!

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

Show that: . $\frac{DA}{XB} \:=\: \frac{AX}{BY} \:=\: \frac{XD}{XY}$

Code:

                X
    A o - - - - o - - - - - - - - - o B
      |      β *    *   α           |
      |       *  90°    *           |
      |      *              *       |
      |     *                   * β |
      |    *                        o Y
      |   *                         |
      |α *                          |
      | *                           |
      |*                            |
    D o - - - - - - - - - - - - - - o C

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

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

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

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

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

**hongvo** · April 28th 2010, 03:58 PM

Hi Soroban

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

Cheers, hv

**imcnalty** · May 5th 2010, 03:02 PM

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

**Soroban** · May 5th 2010, 08:37 PM

Hello, imcnalty!

Who is "sophie"?

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

Find the lengths of the sides of $\Delta DXY.$

Code:

          15    X        48
    A o - - - - o - - - - - - - - - o B
      |        *    *               |
      |       *  90°    *           | b
      |      *              *       |
    a |     *                   * β |
      |    *                        o Y
      |   *                         |
      |α *                          | 16
      | *                           |
      |*                            |
    D o - - - - - - - - - - - - - - o C
                    63

Draw $DY.\quad DC = 63.$

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

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

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

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

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

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

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

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

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

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

**katepark72** · May 6th 2010, 02:25 AM

hi can you help me with this one to??

DX= 429
DY= 845!!
hurry please

thankyou soooo muchhh xxxx

**hongvo** · May 6th 2010, 05:42 AM

Hi Soroban

I've used the Pythagoras triangle (Triads) to get the answers. Is it still correct? The answers are the same in the end.

Cheers, hv

**MathsWiz** · May 10th 2010, 03:23 AM

Sup katepark , are you in year 9 or 10? I suppose you're doing the Maths Challenge too yeah?

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