0
$\begingroup$

Find the angles of an inscribed trapezoid (in a circle) $ABCD$ ($AB||CD$) if $\angle ABD = 63^\circ$.


Any trapezium in a circle is an isosceles trapezium, so $AD = BC$, thus $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AD} = \newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BC} = 2\cdot63^\circ = 126^\circ$. I've tried to calculate some angles if $P = AC$ $\cap$ $BD$ : $\angle APD = 126^\circ$ and $\angle APB = 54^\circ$. It seems useless and I think that there's a missing information. Is is possible to solve the problem?

$\endgroup$
1
$\begingroup$

It is not possible to solve the problem with the given information. Any isosceles trapezoid can be inscribed in a circle. Then let's start with some given $AB$ segment, and we draw a line from $A$ and one from $B$ at the given angle, that will intersect at point $P$ in your figure. Now choose any point $D$ on the extension of $BP$, away from $B$, on the same side as $P$, then draw a parallel to $AB$. This will intersect the extension of $AP$ in $C$. You can immediately see that this is an isosceles trapezoid, that can be inscribed in a circle. Now choose a point $D'$, find $C'$, similarly to the procedure above. Once again $ABC'D'$ is an isosceles trapezoid, which can be inscribed in a circle, but $\angle BAD\ne\angle BAD'$. Therefore you cannot solve the original problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.