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Question Number 88301 by byaw last updated on 09/Apr/20
A circle touches the four sides  of quadrilateral ABCD. Show/prove  that AB+CD=BC+DA.  Please help.
$${A}\:{circle}\:{touches}\:{the}\:{four}\:{sides} \\ $$$${of}\:{quadrilateral}\:{ABCD}.\:\mathrm{Show}/\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{A}{B}+{CD}={BC}+{DA}. \\ $$$$\mathrm{Please}\:\mathrm{help}. \\ $$
Commented by mr W last updated on 09/Apr/20
Commented by mr W last updated on 10/Apr/20
AB+CD=AE+EB+DG+GC  =AH+BF+HD+FC  =AH+HD+BF+FC  =AD+BC  =BC+DA
$${AB}+{CD}={AE}+{EB}+{DG}+{GC} \\ $$$$={AH}+{BF}+{HD}+{FC} \\ $$$$={AH}+{HD}+{BF}+{FC} \\ $$$$={AD}+{BC} \\ $$$$={BC}+{DA} \\ $$

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