Question

prove that \\( \angle abc \\) is a right angle.
given: \\( \overline{ad} \cong \overline{bd} \cong \overline{cd} \\)
prove: \\( m\angle abc = 90 \\)
isosceles triangle theorem
same-side interior angles postulate
triangle angle-sum theorem
algebraic simplification
substitution
alternate interior angles theorem
triangle exterior angle theorem
\\( \overline{ad} \cong \overline{bd} \cong \overline{cd} \\)
\\( \triangle adb \\) and \\( \triangle cdb \\) are isosceles.
\\( m\angle dab = x \\)
\\( m\angle dba = x \\)
\\( m\angle cdb = 2x \\)
\\( m\angle bcd + m\angle dbc + m\angle cdb = 180 \\)
\\( m\angle bcd = m\angle dbc \\)
\\( 2\cdot m\angle bcd + 2x = 180 \\)

Explanation:

Step1: Use Isosceles Triangle Theorem

Since \(AD\cong BD\), in \(\triangle ADB\), by the Isosceles Triangle Theorem (if two sides of a triangle are congruent, then the angles opposite those sides are congruent), \(m\angle DAB = m\angle DBA=x\).

Step2: Use Triangle Exterior Angle Theorem

For \(\triangle ADB\), \(\angle CDB\) is an exterior angle. By the Triangle Exterior Angle Theorem (\(m\angle CDB=m\angle DAB + m\angle DBA\)), so \(m\angle CDB = 2x\).

Step3: Use Isosceles Triangle Theorem again

Since \(BD\cong CD\), in \(\triangle CDB\), by the Isosceles Triangle Theorem, \(m\angle BCD=m\angle DBC\).

Step4: Use Triangle Angle - Sum Theorem

In \(\triangle CDB\), by the Triangle Angle - Sum Theorem (\(m\angle BCD + m\angle DBC+m\angle CDB = 180^{\circ}\)). Substitute \(m\angle BCD=m\angle DBC\) and \(m\angle CDB = 2x\) into the equation, we get \(2\cdot m\angle BCD+2x = 180\).

Step5: Find \(m\angle ABC\)

\(m\angle ABC=m\angle ABD + m\angle DBC\). From \(2\cdot m\angle BCD+2x = 180\), we can rewrite it as \(m\angle BCD + x=90\). And \(m\angle ABC=x + m\angle DBC\), since \(m\angle BCD=m\angle DBC\), so \(m\angle ABC = 90^{\circ}\).

Answer:

\(m\angle ABC = 90^{\circ}\), so \(\angle ABC\) is a right - angle.