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Problem 545

AMC 10/12, early questions
Geometry Difficulty 3.4 Multiple choice CEMC Cayley · Canada · 2026

In the diagram, ABCDABCD is a
rectangle and △ACE\triangle ACE is
right-angled at CC.

If ∠BAC=42°\angle BAC=42\degree, the
measure of ∠DCE\angle DCE is

Pick one

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Official solution

Solution 1:

In △ABC\triangle ABC, $∠ACB=180°−∠ABC−∠BAC=180°−90°−42°=48°$.\$\angle ACB=180\degree-\angle ABC-\angle BAC=180\degree-90\degree-42\degree=48\degree\$.

Since ∠BCD=90°\angle BCD=90\degree, then
$∠ACD=∠BCD−∠ACB=90°−48°=42°$.\$\angle ACD=\angle BCD-\angle ACB=90\degree-48\degree=42\degree\$.

Since ∠ACE=90°\angle ACE=90\degree, then
$∠DCE=∠ACE−∠ACD=90°−42°=48°$.\$\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree\$.

Solution 2:

Since $∠ABC=∠BCD=90°\$\angle ABC=\angle BCD=90\degree,then, then AB$ is
parallel to DCDC.

Thus, $∠ACD=∠\$\angle ACD=\angle BAC
=42°=42\degreebyaparallellinestheorem( by a parallel lines theorem (Z$ pattern, alternate interior
angles).

Since ∠ACE=90°\angle ACE=90\degree, then
$∠DCE=∠ACE−∠ACD=90°−42°=48°$.\$\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree\$.

Solution 3:

Since ∠BAD=90°\angle BAD=90\degree, then
$∠CAD=∠BAD−∠BAC=90°−42°=48°$.\$\angle CAD=\angle BAD-\angle BAC=90\degree-42\degree=48\degree\$.

In △ADC\triangle ADC, $∠ACD=180°−∠ADC−∠CAD=180°−90°−48°=42°$.\$\angle ACD=180\degree-\angle ADC-\angle CAD=180\degree-90\degree-48\degree=42\degree\$.

Since ∠ACE=90°\angle ACE=90\degree, then
$∠DCE=∠ACE−∠ACD=90°−42°=48°$.\$\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree\$.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.