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14. Let $\displaystyle ABC$ be a right triangle: .$\displaystyle \angle C = 90^o.$
A circle $\displaystyle O$ has side $\displaystyle AC$ as its diameter, meets hypotenuse $\displaystyle AB$ at $\displaystyle E.$
A tangent line to $\displaystyle O$ at $\displaystyle E$ meets side $\displaystyle BC$ at point $\displaystyle D.$
Prove that $\displaystyle \Delta BDE$ is isosceles. Code:
F
o
\
θ'\
A o * \
| θ * *\ E
| θ o
| * * *
| * \ θ ' *
| * *\ *
O * * \ *
| * \ *
| \ *
| * \ *
| * \ *
| * \ θ' *
C o * - - - - - - - o - - - - - - - - - - - - - - o B
D
Draw radius $\displaystyle OE.$
Since $\displaystyle OA = OE,\;\Delta AOE$ is isosceles.
. . $\displaystyle \angle OAE = \angle OEA = \theta$
Since $\displaystyle D{E}F$ is tangent at $\displaystyle E,\;\angle OEF = 90^o.$
. . Hence, $\displaystyle \angle OEA$ and $\displaystyle \angle AEF$ are complementary.
. . Let $\displaystyle \angle AEF = \theta'$
$\displaystyle \angle AEF$ and $\displaystyle \angle DEB$ are vertical angles: .$\displaystyle {\color{blue}\angle DEB = \theta'}$
In $\displaystyle \Delta ABC,\;\angle CAB= \theta $ and $\displaystyle \angle ABC$ are complementary.
. . Hence: .$\displaystyle {\color{blue}\angle ABC = \theta'}$
In $\displaystyle \Delta BDE\!:\;\angle DEB = \angle EBD = \theta'$
Therefore, $\displaystyle \Delta BDE$ is isosceles.