Question

$A\equiv (0,b),B=(0,0)$ and $C=(a,0)$ are the vertices of $△ABC.D,E,F$ are the mid-points of the sides $BC,CA$ and $AB$ respectively. If $a^2+b^2=20$ then

• A. $(AD{)}^2=9$
• B. $(BE{)}^2=4$
• C. $(AD{)}^2+(CF{)}^2=25$
• D. $(AD{)}^2+(CF{)}^2=(BE{)}^2$

Answer 1

Answer: (C), (D)

The correct options are
C $(AD{)}^2+(CF{)}^2=25$
D $(AD{)}^2+(CF{)}^2=(BE{)}^2$

As $A\equiv (0,b),B\equiv (0,0)$ and $C\equiv (a,0)$
Then $D\equiv (\frac{a}{2},0),E\equiv (\frac{a}{2},\frac{b}{2})$ and $F\equiv (0,\frac{b}{2})$
Using $a^2+b^2=20$, we have
${\left(AD\right)}^2={(\frac{a}{2}-0)}^2+{(0-b)}^2=\frac{a^2}{4}+b^{2}Similary{\left(CF\right)}^2{\left(AD\right)}^2+{\left(CF\right)}^2={(\frac{a}{2}-0)}^2+{(0-b)}^2+{(0-a)}^2+{(\frac{b}{2}-0)}^2=\frac{5(a^2+b^2)}{4}=25={\left(BE\right)}^2$