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CEMC Cayley, 2026

Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, $ABCD$ is a
rectangle and $\triangle ACE$ is
right-angled at $C$ .

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

Pick one

Solution

Solution 1:

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

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

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

Solution 2:

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

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

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

Solution 3:

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

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

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

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