1. ## Geometric Proof

1. The vertex E of a square EFGH is inside a square ABCD. The vertices F, G and H are outside the square ABCD. The side EF meets the side CD at X and the side EH meets the side AD at Y. If EX = EY, prove that E lies on BD.

2. Two circles C1 and C2 meet at the points A and B. The tangent to C1 at A meets C2 at P. A point Q inside C1 lies on the circumference of C2. When produced, BQ meets C1 at S and PA produced at T. Prove that AS is parallel to PQ.

Cheers on helping out

2. ## Re: Geometric Proof

Hello, john128!

1. The vertex E of a square EFGH is inside a square ABCD.
The vertices F, G and H are outside the square ABCD.
The side EF meets the side CD at X and the side EH meets the side AD at Y.
If EX = EY, prove that E lies on BD.

Code:
                A               B
H   o---------------o
o   |               |
*   Y*               |
*     |θ  *           |
*     Z+ - - - oE      |
*       |      *:       |
*        |     * :       |
G o         |    *θ :       |
*   D o---*---+-------o C
*    *X   W
o
F
Draw $\displaystyle EW \perp CD.$
Draw $\displaystyle EZ \perp AD.$
Let $\displaystyle \angle EXW \,=\,\theta \,=\,\angle EYZ.$

In $\displaystyle \Delta EW\!X\!:\;EW \,=\,EX\sin\theta$
In $\displaystyle \Delta EZY\!:\;EZ \,=\,EY\sin\theta$

Since $\displaystyle EX = EY$, then $\displaystyle EW = EZ.$

Hence, $\displaystyle E$ is equidistant from sides $\displaystyle AD$ and $\displaystyle CD.$
That is, $\displaystyle E$ is on the bisector of $\displaystyle \angle C.$

Therefore, $\displaystyle E$ is on the diagonal $\displaystyle BD.$

4. ## Re: Geometric Proof

Originally Posted by john128
2. Two circles C1 and C2 meet at the points A and B. The tangent to C1 at A meets C2 at P. A point Q inside C1 lies on the circumference of C2. When produced, BQ meets C1 at S and PA produced at T. Prove that AS is parallel to PQ.

In the circle $\displaystyle C_2, \ \widehat{BQP}=\widehat{BAP}=\frac{arc(BP)}{2}$.
But in the circle $\displaystyle C_1, \ \widehat{BAP}=\frac{arc(AB)}{2}$ and $\displaystyle \widehat{BSA}=\frac{arc(AB)}{2}$.
Then $\displaystyle \widehat{BQP}=\widehat{BSA}\Rightarrow PQ \parallel AS$