# deductive geometry (1)

• Oct 30th 2009, 02:50 AM
thereddevils
deductive geometry (1)
Points A and B are on the side XY of triangle XYZ with XA=AB=BY . Points C and D are on the sides YZ and XZ respectively such that ABCD is a rhombus . Prove that angle XYZ is 90 degree .
• Oct 30th 2009, 08:28 AM
Soroban
Hello, thereddevils!

Please check the wording of the problem.
As stated, the conclusion cannot be true.

Quote:

Points $\displaystyle A$ and $\displaystyle B$ are on the side $\displaystyle XY$ of $\displaystyle \Delta XYZ$ with $\displaystyle XA=AB=BY.$
Points $\displaystyle C$ and $\displaystyle D$ are on the sides $\displaystyle YZ$ and $\displaystyle XZ$ resp. such that $\displaystyle ABCD$ is a rhombus.
Prove that $\displaystyle \angle XYZ \,=\,90^o.$

Code:

    X o       |\       | \       |  \       |  \       |    \     A o    \       | *    \       |  *  \       |    *  \       |      * \       |        *\     B o          o D       | *        |\       |  *      | \       |    *    |  \       |      *  |  \       |        * |    \     Y o - - - - - o - - o Z                   C
Suppose $\displaystyle \angle XYZ = 90^o.$

We are given: .$\displaystyle BY = AB$

Since $\displaystyle ABCD$ is a rhombus: .$\displaystyle BC = AB.$

Hence: .$\displaystyle BY = BC.$

Then we have right triangle $\displaystyle BYC$ in which side $\displaystyle BY$
. . and hypotenuse $\displaystyle BC$ are equal. ??

• Oct 30th 2009, 10:07 AM
EekNah90
not sure.. perhaps they ask u to prove angle XZY is 90
• Oct 30th 2009, 10:35 PM
thereddevils
Quote:

Originally Posted by Soroban
Hello, thereddevils!

Please check the wording of the problem.
As stated, the conclusion cannot be true.

Code:

    X o       |\       | \       |  \       |  \       |    \     A o    \       | *    \       |  *  \       |    *  \       |      * \       |        *\     B o          o D       | *        |\       |  *      | \       |    *    |  \       |      *  |  \       |        * |    \     Y o - - - - - o - - o Z                   C
Suppose $\displaystyle \angle XYZ = 90^o.$

We are given: .$\displaystyle BY = AB$

Since $\displaystyle ABCD$ is a rhombus: .$\displaystyle BC = AB.$

Hence: .$\displaystyle BY = BC.$

Then we have right triangle $\displaystyle BYC$ in which side $\displaystyle BY$
. . and hypotenuse $\displaystyle BC$ are equal. ??

Sorry Soroban , it should be prove that angle XZY = 90 .
• Oct 31st 2009, 08:41 AM
Soroban
Hello, thereddevils!

Quote:

It should be: prove that $\displaystyle \angle XZY \,=\, 90^o$ . . . . Ah, got it!

Code:

    X o       |\       | \       |  \       |  \       |    \       |    o A       |    / \       |  / θ \       |  /    \       | /      \       |/        \     D o          o B       |\        / \       | \      / θ \       |  \    /    \       |  \  /      \       |    \ /        \     Z o - - o - - - - - o Y             C

We have: .$\displaystyle XA \,=\,AB \,=\,BY \,=\,BC\,=\,CD\,=\,DA$

Let $\displaystyle \theta \,=\,\angle CBY$

Since $\displaystyle \Delta CBY$ is isosceles, $\displaystyle \angle BCY = \angle BYC.$
Then: .$\displaystyle \angle BYC + \angle BCY + \theta \:=\:180^o \quad\Rightarrow\quad \boxed{\angle BYC \,=\,90^o - \tfrac{\theta}{2}}$

Since $\displaystyle AD \parallel BC\!:\;\angle BAD = \theta \quad\Rightarrow\quad \angle XAD \,=\,180^o-\theta$

Since $\displaystyle \Delta XAD$ is isosceles: $\displaystyle \angle AXD \,=\,\angle ADX$
Then: .$\displaystyle \angle AXD + \angle ADX + (180^o -\theta) \:=\:180^o \quad\Rightarrow\quad \boxed{\angle AXD \,=\,\tfrac{\theta}{2}}$

In $\displaystyle \Delta XYZ\!:\;\;\angle X \,=\, \tfrac{\theta}{2},\;\;\angle Y \,=\, 90^o-\tfrac{\theta}{2}$ . . . They are complementary!

Therefore: .$\displaystyle \angle Z \:=\:90^o$