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Question Number 173668 by mr W last updated on 16/Jul/22

Commented by mr W last updated on 16/Jul/22

$p r o v e t h a t M i s t h e m i d p o i n t o f E F .$
	Answered by mr W last updated on 16/Jul/22

	Answered by mr W last updated on 16/Jul/22

	Commented by infinityaction last updated on 16/Jul/22

	Commented by infinityaction last updated on 16/Jul/22

	$A B I ⋍ A E F$
	Commented by Tawa11 last updated on 16/Jul/22

	$Great sir$
	Answered by mindispower last updated on 16/Jul/22
	$let (B,BA,BD) repere B(0,0);A(1,0);D(0,1) E(1,1);F(a,b) ((z_c −z_A )/(z_F −z_A ))=i⇒Z_c =1+b−i(a−1) C(1+b,1−a) equation of (AH) AM^→ .BC^→ =0⇒ (((x−1)),(y) ). (((1+b)),((1−a)) )=0 (1+b)(x−1)+(1−a)y=0 equation of (BF) (b−1)(x−1)+(1−a)(y−1)=0 M(x,y)∈(EF)∩(AH) { (((1+b)(x−1)+(1−a)y=0)),(((b−1)(x−1)+(1−a)(y−1)=0)) :} ⇒2(x−1)=a−1⇒x=((1+a)/2) y=((1+b)/(a−1))(((a−1)/2))=((1+b)/2) M(((1+a)/2),((1+b)/2))=(((x_E +x_F )/2),((y_E +y_F )/2))$
	$l e t (B, B A, B D) r e p e r e$ $B (0, 0); A (1, 0); D (0, 1)$ $E (1, 1); F (a, b)$ $\frac{z_{c} - z_{A}}{z_{F} - z_{A}} = i \Rightarrow Z_{c} = 1 + b - i (a - 1)$ $C (1 + b, 1 - a)$ $e q u a t i o n o f (A H)$ $A \vec{M} . B \vec{C} = 0 \Rightarrow (\begin{matrix} x - 1 \\ y \end{matrix}) . (\begin{matrix} 1 + b \\ 1 - a \end{matrix}) = 0$ $(1 + b) (x - 1) + (1 - a) y = 0$ $e q u a t i o n o f (B F)$ $(b - 1) (x - 1) + (1 - a) (y - 1) = 0$ $M (x, y) \in (E F) \cap (A H)$ ${\begin{cases} (1 + b) (x - 1) + (1 - a) y = 0 \\ (b - 1) (x - 1) + (1 - a) (y - 1) = 0 \end{cases}$ $\Rightarrow 2 (x - 1) = a - 1 \Rightarrow x = \frac{1 + a}{2}$ $y = \frac{1 + b}{a - 1} (\frac{a - 1}{2}) = \frac{1 + b}{2}$ $M (\frac{1 + a}{2}, \frac{1 + b}{2}) = (\frac{x_{E} + x_{F}}{2}, \frac{y_{E} + y_{F}}{2})$
	Commented by Tawa11 last updated on 16/Jul/22

	$Great sir$
	Commented by mindispower last updated on 17/Jul/22

	$w i t h e P l e a s u r h a v e a n i c e D a y$
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