Question Number 105014 by mathocean1 last updated on 25/Jul/20
$$\left.\mathrm{1}\right){Determinate}\:\alpha\:{then}\:{show}\:{that}\:{I}\:{belong}\:{to}\:{the}\:{circle} \\ $$$$\left.\mathrm{2}\right)\:{show}\:{that}\:\left({BF}\right)\bot\left({DI}\right).\: \\ $$$$ \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 25/Jul/20
$$\mathrm{BF}\bot\mathrm{DI}\:,\mathrm{have}\:\mathrm{not}\:\mathrm{BF}\bot\mathrm{DE}\:\mathrm{you}\:\mathrm{look}\:\mathrm{at} \\ $$$$\mathrm{once}\:\mathrm{again}\:\mathrm{your}\:\mathrm{figure}? \\ $$
Commented by mathocean1 last updated on 25/Jul/20
$${yes}\:{sir}\:{it}\:{was}\:{an}\:{error}. \\ $$$${Show}\:{that}\:{BF}\bot{DI} \\ $$
Commented by mr W last updated on 25/Jul/20
$${see}\:{Q}\mathrm{104998} \\ $$
Commented by 1549442205PVT last updated on 28/Jul/20
$$\mathrm{It}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{becau}\:\mathrm{se}\:\mathrm{we}\:\mathrm{proved}\:\mathrm{that}\:\mathrm{I} \\ $$$$\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}.\mathrm{Then}\:\widehat {\mathrm{BID}}=\mathrm{90}°\:\mathrm{since} \\ $$$$\mathrm{BD}\:\mathrm{is}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$