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Question-198421




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

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