Question Number 198421 by mr W last updated on 19/Oct/23
Commented by mr W last updated on 19/Oct/23
$${given}\:{that}\:{the}\:{midpoints}\:{of}\:{the}\:{sides} \\ $$$${of}\:{a}\:{convex}\:{quadrilaterial}\:{are} \\ $$$${concyclic}.\:{if}\:{the}\:{lengthes}\:{of}\:{three}\:{of} \\ $$$${its}\:{sides}\:{are}\:{a},\:{b},\:{c}\:{respectively},\:{find} \\ $$$${the}\:{length}\:{of}\:{its}\:{fourth}\:{side}. \\ $$
Commented by som(math1967) last updated on 20/Oct/23
$${Mid}\:{pt}\:{form}\:{a}\:{rectangle}\:\left({cyclic}\:{parelallogram}\right) \\ $$$${so}\:{AB}={AD}\:.\:{am}\:{i}\:{correct}\:{sir}\:? \\ $$
Commented by mr W last updated on 20/Oct/23
$${yes} \\ $$
Answered by AST last updated on 20/Oct/23
$${Diagonals}\:{intersect}\:{at}\:{right}\:{angles} \\ $$$${Pythagoras}\:{theorem}\:{for}\:{the}\:{sides}\:{give}: \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} ={b}^{\mathrm{2}} +{d}^{\mathrm{2}} \Rightarrow{d}=\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 20/Oct/23
$${thanks}\:{for}\:{the}\:{right}\:{answer}! \\ $$
Answered by mr W last updated on 20/Oct/23
Commented by mr W last updated on 20/Oct/23
$${EH}={FG}=\frac{{BD}}{\mathrm{2}} \\ $$$${EH}//{FG}//{BD} \\ $$$${EF}={HG}=\frac{{AC}}{\mathrm{2}} \\ $$$${EF}//{HG}//{AC} \\ $$$$\Rightarrow{EFGH}\:{is}\:{parallelogram} \\ $$$${since}\:{EFGH}\:{are}\:{concyclic} \\ $$$$\Rightarrow{EFGH}\:{is}\:{rectangle} \\ $$$$\Rightarrow{AC}\bot{BD} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} ={AK}^{\mathrm{2}} +{BK}^{\mathrm{2}} +{CK}^{\mathrm{2}} +{DK}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{d}^{\mathrm{2}} ={CK}^{\mathrm{2}} +{BK}^{\mathrm{2}} +{AK}^{\mathrm{2}} +{DK}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{c}^{\mathrm{2}} ={b}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$$\Rightarrow{d}=\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$