Question

In $△ABC,AX\perp BC$ and $Y$ is middle point of $BC$. Then prove that
(i) ${AB}^2={AY}^2+\frac{{BC}^2}{4}-BC.XY$
(i) ${AC}^2={AY}^2+\frac{{BC}^2}{4}+BC.XY$

Answer 1

(1) In $△ABC$

$(AB{)}^2=(AX{)}^2+(BX{)}^2$

$=(AX{)}^2+(BX-XY{)}^2$

$=(AX{)}^2+(BX{)}^2+(XY{)}^2-2BY.XY$

$=(AX{)}^2+{\left[\frac{BC}{2}\right]}^2+(XY{)}^2-2\frac{BC}{2}.XY$

$=(AX{)}^2+(XY{)}^2+\frac{(BC{)}^2}{4}-BC.XY$

$=(AY{)}^2+\frac{(BC{)}^2}{4}-BC.XY$ $\left[In△AXY\right(AY{)}^2=(AX{)}^2+(XY{)}^2]$

(2) In $△AXC$

$(AC{)}^2=(AX{)}^2+(XC{)}^2$

$=(AX{)}^2+(XY+YC{)}^2$

$=(AX{)}^2+(XY{)}^2+(YC{)}^2+2XY,YC$

$=(AX{)}^2+(XY{)}^2+{\left[\frac{BC}{2}\right]}^2+2\frac{BC}{2}.XY$

$=(AY{)}^2+\frac{(BC{)}^2}{4}+BC.XY$ $\left[In△AXY\right(AY{)}^2=(AX{)}^2+(XY{)}^2]$