IV- (6 points)
In an orthonormal system of axes( $x^{'} Ox, y^{'} Oy)$ , consider the points $A (2; 0), B (0; 4)$ and $E (- 4; 0)$ .
Let (d) be the line with equation $y = - 2 x + 4$ .
1) Plot the points $A, B$ and $E$ .
2) Verify that $A$ and $B$ are two points on (d), then draw (d).
3) Let ( $d^{'})$ be the line passing through $E$ and perpendicular to (d).
Verify that $y = \frac{1}{2} x + 2$ is the equation of $(d^{'})$ .
4) The line (d') intersects ( $y^{'} Oy)$ at $H (0; 2)$ and intersects (d) at $F$ .
a. Verify that the coordinates of $F$ are $(\frac{4}{5}; \frac{12}{5})$ .
b. Show that $H$ is the orthocenter of triangle $EAB$ .
c. Prove that (AH) is perpendicular to (EB).
5) The line (AH) intersects (EB) at G.
a. Prove that the four points E, G, F and $A$ are on the same circle (C) with diameter to be determined.
b. Calculate the radius of (C).

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5b) GE = $\sqrt{(- 4 - 2)^{2} + (0 - 0)^{2}}$ =6, EF = $\sqrt{{(\frac{4}{5} - (- 4))}^{2} + {(\frac{12}{5} - 0)}^{2}}$ = $\frac{40}{\sqrt{5}}$ , Circle radius = $\frac{1}{2} \sqrt{40^{2} + (\frac{40}{\sqrt{5}})^{2}}$ = $\frac{40}{\sqrt{5}}$

Steps

Step 1 ：1) A(2;0), B(0;4), E(-4;0)

Step 2 ：2) y=-2x+4, A y(-2(2)+4)=0, B y(-2(0)+4)=4, AB on (d)

Step 3 ：3) y=\frac{1}{2}x+2, \frac{1}{2}(-4)+2=0, E on $(d^{'})$ , $(d^{'})$ is perpendicular to (d)

Step 4 ：4a) F coordinates $(\frac{4}{5}; \frac{12}{5})$ , calculation: x = $\frac{y - 2}{\frac{1}{2}}$ , substitute y into (d): x = $\frac{- 2 x + 4 - 2}{\frac{1}{2}}$ , solve the equation for x: x = $\frac{4}{5}$

Step 5 ：4b) H=(0;2), HF (dB\) is perpendicular to EB, AF (dA\) is perpendicular to EA, HA and HE form a right angle

Step 6 ：4c) AH * EB = 2 * (-4) + (0 * 0) = -8

Step 7 ：5a) G is the intersection of AH and EB, four points E, G, F and A are on the same circle with diameter EG

Step 8 ：5b) GE = $\sqrt{(- 4 - 2)^{2} + (0 - 0)^{2}}$ =6, EF = $\sqrt{{(\frac{4}{5} - (- 4))}^{2} + {(\frac{12}{5} - 0)}^{2}}$ = $\frac{40}{\sqrt{5}}$ , Circle radius = $\frac{1}{2} \sqrt{40^{2} + (\frac{40}{\sqrt{5}})^{2}}$ = $\frac{40}{\sqrt{5}}$

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