deductive geometry (1)

thereddevils · #1 $Old$ October 30th, 2009, 03:50 AM

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 .

Soroban · #2 $Old$ October 30th, 2009, 09:28 AM

Hello, thereddevils!

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

Quote:

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. ??

EekNah90 · #3 $Old$ October 30th, 2009, 11:07 AM

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

thereddevils · #4 $Old$ October 30th, 2009, 11:35 PM

Quote:

Originally Posted by Soroban $View Post$

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 .

Soroban · #5 $Old$ October 31st, 2009, 09:41 AM

Hello, thereddevils!

Quote:

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