Question

fill out the blanks
hint for ( mangle bca )

Explanation:

Step1: Find \(m\angle ABC\)

Since \(AB\perp BE\), \(m\angle ABC = 90^{\circ}\)

Step2: Find \(m\angle BCA\)

In \(\triangle ABC\), \(\angle A = 45^{\circ}\), \(\angle ABC=90^{\circ}\). Using the angle - sum property of a triangle (\(\angle A+\angle ABC+\angle BCA = 180^{\circ}\)), we have \(45^{\circ}+90^{\circ}+\angle BCA=180^{\circ}\). Then \(\angle BCA=180^{\circ}-(45^{\circ} + 90^{\circ})=45^{\circ}\)

Step3: Find \(m\angle BCD\)

\(\angle BCA\) and \(\angle BCD\) are supplementary (\(\angle BCA+\angle BCD = 180^{\circ}\)). Since \(\angle BCA = 45^{\circ}\), then \(\angle BCD=180^{\circ}-45^{\circ}=135^{\circ}\)

Step4: Find \(m\angle ACB + m\angle BCD\)

\(m\angle ACB + m\angle BCD=45^{\circ}+135^{\circ}=180^{\circ}\)

Step5: Find \(m\angle FBE + m\angle CBA\)

\(m\angle FBE = 90^{\circ}\), \(m\angle CBA = 90^{\circ}\), so \(m\angle FBE + m\angle CBA=90^{\circ}+90^{\circ}=180^{\circ}\)

Step6: Find \(m\angle DCG\)

\(\angle DCG\) and \(\angle BCD\) are vertical angles. Since \(m\angle BCD = 135^{\circ}\), then \(m\angle DCG=135^{\circ}\)

Answer:

	Measure
\(m\angle BCA\)	\(45^{\circ}\)
\(m\angle BCD\)	\(135^{\circ}\)
\(m\angle ACB + m\angle BCD\)	\(180^{\circ}\)
\(m\angle FBE + m\angle CBA\)	\(180^{\circ}\)
\(m\angle DCG\)	\(135^{\circ}\)