1. ## 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 .

2. Hello, thereddevils!

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

Points $A$ and $B$ are on the side $XY$ of $\Delta XYZ$ with $XA=AB=BY.$
Points $C$ and $D$ are on the sides $YZ$ and $XZ$ resp. such that $ABCD$ is a rhombus.
Prove that $\angle XYZ \,=\,90^o.$
Code:
    X o
|\
| \
|  \
|   \
|    \
A o     \
| *    \
|   *   \
|     *  \
|       * \
|         *\
B o           o D
| *         |\
|   *       | \
|     *     |  \
|       *   |   \
|         * |    \
Y o - - - - - o - - o Z
C
Suppose $\angle XYZ = 90^o.$

We are given: . $BY = AB$

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

Hence: . $BY = BC.$

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

3. not sure.. perhaps they ask u to prove angle XZY is 90

4. 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 $\angle XYZ = 90^o.$

We are given: . $BY = AB$

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

Hence: . $BY = BC.$

Then we have right triangle $BYC$ in which side $BY$
. . and hypotenuse $BC$ are equal. ??
Sorry Soroban , it should be prove that angle XZY = 90 .

5. Hello, thereddevils!

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

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

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

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

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

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

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

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

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