1974 AMC 12 Problem 5

Attempt Problem 5 of the 1974 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1974 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

Given a quadrilateral ABCDABCD inscribed in a circle with side ABAB extended beyond BB to point E,E, if ∠BAD=92∘\angle BAD=92^\circ and ∠ADC=68∘,\angle ADC=68^\circ, find ∠EBC.\angle EBC.
66∘66^\circ
68∘68^\circ
70∘70^\circ
88∘88^\circ
92∘92^\circ
Answer: B
Concepts:cyclic quadrilateralangle chasing
Small hint:
Opposite angles of a cyclic quadrilateral are supplementary
Big hint:
The desired exterior angle is also supplementary to ∠ABC\angle ABC
Solution:
Because ABCDABCD is cyclic, ∠ABC+∠ADC=180∘.\angle ABC+\angle ADC=180^\circ. Since BEBE extends BA,BA, ∠EBC+∠ABC=180∘\angle EBC+\angle ABC=180^\circ as well. Hence ∠EBC=∠ADC=68∘. \angle EBC=\angle ADC=68^\circ. The given 92∘92^\circ angle is unnecessary.
Therefore, the correct answer is B.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.