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 .
Hello, thereddevils!
Please check the wording of the problem.
As stated, the conclusion cannot be true.
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 $\displaystyle \angle XYZ \,=\,90^o.$Suppose $\displaystyle \angle XYZ = 90^o.$Code:X o |\ | \ | \ | \ | \ A o \ | * \ | * \ | * \ | * \ | *\ B o o D | * |\ | * | \ | * | \ | * | \ | * | \ Y o - - - - - o - - o Z C
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. ??
Hello, thereddevils!
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: .$\displaystyle \angle Z \:=\:90^o$