QUESTION IMAGE
Question
m∠bcd =
m∠ade =
m∠abd =
m∠aeb =
m∠cbe =
m∠dea =
Step1: Identify rectangle properties
In rectangle \(ABCD\), all interior angles are \(90^\circ\), and diagonals are equal and bisect each other (\(AE=DE=BE=CE\)).
Step2: Solve \(m\angle BCD\)
All rectangle angles are \(90^\circ\).
\(m\angle BCD = 90^\circ\)
Step3: Solve \(m\angle ABD\)
\(AB \parallel DC\), so alternate interior angles are equal: \(m\angle ABD = m\angle BDC = 16^\circ\)
Step4: Solve \(m\angle CBE\)
\(m\angle ABC = 90^\circ\), so subtract \(m\angle ABD\).
\(m\angle CBE = 90^\circ - 16^\circ = 74^\circ\)
Step5: Solve \(m\angle ADE\)
\(AE=DE\), so \(\triangle ADE\) is isosceles. \(m\angle ADE = m\angle DAE\). First find \(m\angle ADC = 90^\circ\), so \(m\angle ADE = 90^\circ - 16^\circ = 74^\circ\)
Step6: Solve \(m\angle AEB\)
Use exterior angle of \(\triangle AED\): \(m\angle AEB = m\angle ADE + m\angle DAE\)
\(m\angle AEB = 74^\circ + 74^\circ = 148^\circ\)
Step7: Solve \(m\angle DEA\)
Linear pair with \(m\angle AEB\): subtract from \(180^\circ\).
\(m\angle DEA = 180^\circ - 148^\circ = 32^\circ\)
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- \(m\angle BCD = 90^\circ\)
- \(m\angle ABD = 16^\circ\)
- \(m\angle CBE = 74^\circ\)
- \(m\angle ADE = 74^\circ\)
- \(m\angle AEB = 148^\circ\)
- \(m\angle DEA = 32^\circ\)