Two Quite simple geometry questions

[Jaspworld]

The following questions are probably pretty simple but I don't get them.

Question 1
Opposite sides of a cuclic quadrilateral ABCD are extended to meet at E. Prove that triangle EAC is similar to triangle EDB.

Question 2
In the diagram, tangent PT and the chord AB have equal lengths. Prove that B divided AP in the golden ratio.

 

[soroban]

Hello, Jaspworld!

1) Opposite sides of a cuclic quadrilateral ABCD are extended to meet at E.
Prove that triangle EAC is similar to triangle EDB.

         A    * * *  B
          * - - - - - * - - - - - - - - - - * E
        *               *               *
       *                 *          *
                                *
      *                   * *
      *                 * *C
      *             *     *
                *
       *    *            *
        *               *
       D  *           *
              * * *

Draw AC, AD, BC, and BD.

\(\displaystyle \angle EAC\) and \(\displaystyle \angle EDB\) both subtend arc BC.
. . Hence: \(\displaystyle \:\angle EAC\,=\,\angle EDB\)

Obviously, \(\displaystyle \angle E\,=\,\angle E\)

Hence: \(\displaystyle \:\angle ACE\,=\,\angle DBE\)

Therefore: \(\displaystyle \:\Delta EAC\,\sim\,\Delta EDB\;\) (a.a.a.)

2) In the diagram, tangent PT and the chord AB have equal lengths.
Prove that B divides AP in the golden ratio.

              * * *
          *           *
        *               *
       *                 *
                x         B    1
    A o   o   o   o   o   o   o   o P
      *                   *     o
      *                   *   o
                            o x
       *                 *o
        *               o
          *           * T
              * * *

Let \(\displaystyle AB\,=\,PT\,=\,x\) and let \(\displaystyle BP\,=\,1\)

Theorem: If a tangent and a secant are drawn from an external point, the tangent
is the mean proportional between the extrenal segment of the secant and the entire secant.

Hence, we have: \(\displaystyle \:x^2 \:=\:1(x\,+\,1)\)

. . which is the quadratic: \(\displaystyle \:x^2\,-\,x\,-\,1\:=\:0\)

. . which has roots: \(\displaystyle \:x\:=\:\frac{1\,\pm\,\sqrt{5}}{2}\)

The positive root is: \(\displaystyle \:x\:=\:\frac{1\,+\,\sqrt{5}}{2}\:=\:\phi\)

Therefore, B divides AP in the golden ratio, \(\displaystyle \phi\,:\,1\)