C-08-Q135mediumsingle_mcq

In the proof, $E$ is the second intersection point of side $AD$ with the auxiliary circle through $A,B,C$ . We deduce $\angle AEC = \angle ADC$ , which contradicts—

athe angle sum of a triangle
ban exterior angle being greater than the opposite interior angle
cthe law of sines
dthe law of cosines

ব্যাখ্যা

Because $\angle AEC$ is an exterior angle of triangle $CDE$ , it must be greater than the opposite interior angle $\angle ADC$ ; the equality $\angle AEC = \angle ADC$ contradicts this exterior-angle inequality, so the fourth vertex must lie on the circle.

In the proof, EEE is the second intersection point of side ADADAD with the auxiliary circle through A,B,CA,B,CA,B,C. We deduce ∠AEC=∠ADC\angle AEC = \angle ADC∠AEC=∠ADC, which contradicts—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof, $E$ is the second intersection point of side $AD$ with the auxiliary circle through $A,B,C$ . We deduce $\angle AEC = \angle ADC$ , which contradicts—