ABC üçgeninde BE CE'ye dik,AD//EC,|AD|=12,|EB|=9,A(ABC)?

Question

*54

Answer 1

$|BF|=x$ ve $|FD|=xk$ olsun.

$|AF|=12-xk$ ve $|FE|=9-x$ olur.

$[FE] \cap [AC]=\{G\}$ olsun.

$|FG|=y$ olsun. $|GE|=9-x-y$ olur.

$\widehat{AFG} \equiv \widehat{GEC}$

$\frac{|AF|}{|FG|}=\frac{|EC|}{|GE|} \\ \frac{12-xk}{y}=\frac{9k}{9-x-y} \\ 108-9xk+x^2k-12x+xyk-12y=9ky \\ 108=9k(x+y)+12(x+y)-xk(x+y)=(x+y)(12-xk+9k) *$

$A(\widehat{ABC})=\frac{|BG|(|AF|+|EC|)}{2}=\frac{(x+y)(12-xk+9k)}{2}$

$*$ eşitliğinden,

$A(\widehat{ABC})=\frac{108}{2}=54$