# Diagram problem

• Apr 28th 2010, 01:28 AM
hongvo
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.
• Apr 28th 2010, 06:00 AM
Soroban
Hello, hongvo!

Quote:

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

• Apr 28th 2010, 03:58 PM
hongvo
diagram problem
Hi Soroban

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

Cheers, hv
• May 5th 2010, 03:02 PM
imcnalty
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
• May 5th 2010, 08:37 PM
Soroban
Hello, imcnalty!

Who is "sophie"?

Quote:

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

• May 6th 2010, 02:25 AM
katepark72
sophie diagram
hi can you help me with this one to??

DX= 429
DY= 845!!