Question Number 78427 by Chi Mes Try last updated on 17/Jan/20
$${please}\:{i}\:{need}\:{it}\:{urgently} \\ $$$$ \\ $$$${show}\:{that}\:{the}\:{midpoint}\:{of}\:{the}\:{hypotenuse} \\ $$$${of}\:{a}\:{right}\:{triangle}\:{is}\:{equidistant}\:{from}\:{its}\:{vertices} \\ $$
Answered by MJS last updated on 17/Jan/20
$$\mathrm{easy}: \\ $$$$\mathrm{any}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{half}\:\mathrm{a}\:\mathrm{rectangle}. \\ $$$$\mathrm{any}\:\mathrm{rectangle}\:\mathrm{has}\:\mathrm{a}\:\mathrm{circumcircle}\:\mathrm{with}\:\mathrm{center} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{distance}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{center} \\ $$
Answered by $@ty@m123 last updated on 17/Jan/20
$${Let}\:\bigtriangleup{ABC}\:{is}\:{right}\:{angled}\:{at}\:{B}. \\ $$$${Let}\:{D}\:{be}\:{the}\:{mid}\:{point}\:{of}\:{AC}. \\ $$$${Draw}\:{DE}\bot{BC}\:\Rightarrow\:{DE}\parallel{AB}. \\ $$$${Converse}\:{of}\:{mid}\:{point}\:{theorem} \\ $$$${implies}\:{that}\:{E}\:{is}\:{mid}\:{point}\:{of}\:{BC}. \\ $$$${Now}\:{in}\:\bigtriangleup{DBE}, \\ $$$${DB}^{\mathrm{2}} ={DE}^{\mathrm{2}} +{EB}^{\mathrm{2}} \\ $$$${DB}^{\mathrm{2}} =\left(\frac{{AB}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{BC}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${DB}^{\mathrm{2}} =\left(\frac{{AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} }{\mathrm{4}}\right)=\frac{{AC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{DB}=\frac{{AC}}{\mathrm{2}} \\ $$$$\Rightarrow{DB}={AD}={DC} \\ $$$${Hence}\:{the}\:{result}. \\ $$
Commented by $@ty@m123 last updated on 17/Jan/20
