Question

drag the tiles to the correct boxes to complete the pairs. match the statements with their values. ( m angle a b c+m angle b a c+m angle a c b ) when ( \triangle a b c ) is an isosceles triangle with ( overline{a b}=overline{a c} ) ( m angle a b c ) when ( m angle b a c=70^{circ} ) and ( \triangle a b c ) is an isosceles triangle with ( overline{a b}=overline{a c} ) ( m angle q p r ) when ( m angle q r p=30^{circ} ) and ( \triangle p q r ) is an isosceles triangle with ( overline{p q}=overline{q r} ) ( m angle b d e ) when ( m angle b a c=45^{circ} ) and points ( d ) and ( e ) are the midpoints of ( overline{a b} ) and ( overline{b c} ), respectively, in ( \triangle a b c )

Explanation:

Step1: Sum of angles in a triangle

By the angle - sum property of a triangle, the sum of the interior angles of any triangle is \(180^{\circ}\). So, \(m\angle ABC + m\angle BAC + m\angle ACB=180^{\circ}\) when \(\triangle ABC\) is an isosceles triangle with \(\overline{AB}=\overline{AC}\).

Step2: Angles in an isosceles triangle \(\triangle ABC\) with \(\overline{AB}=\overline{AC}\)

If \(\overline{AB}=\overline{AC}\), then \(\angle ABC=\angle ACB\). Using the angle - sum property \(m\angle ABC + m\angle BAC + m\angle ACB = 180^{\circ}\), and substituting \(m\angle BAC = 70^{\circ}\) and \(m\angle ABC=m\angle ACB\), we get \(2m\angle ABC+70^{\circ}=180^{\circ}\). Solving for \(m\angle ABC\):

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Step3: Angles in an isosceles triangle \(\triangle PQR\) with \(\overline{PQ}=\overline{QR}\)

If \(\overline{PQ}=\overline{QR}\), then \(\angle QRP=\angle QPR\). Given \(m\angle QRP = 30^{\circ}\), so \(m\angle QPR=30^{\circ}\)

Step4: Mid - segment and parallel lines (Mid - point theorem)

Since \(D\) and \(E\) are mid - points of \(\overline{AB}\) and \(\overline{BC}\) respectively in \(\triangle ABC\), by the mid - point theorem, \(DE\parallel AC\). Then \(\angle BDE=\angle BAC\) (corresponding angles). Given \(m\angle BAC = 45^{\circ}\), so \(m\angle BDE = 45^{\circ}\)

Answer:

\(m\angle ABC + m\angle BAC + m\angle ACB\) (when \(\triangle ABC\) is an isosceles triangle with \(\overline{AB}=\overline{AC}\)) \(\to180^{\circ}\); \(m\angle ABC\) (when \(m\angle BAC = 70^{\circ}\) and \(\triangle ABC\) is an isosceles triangle with \(\overline{AB}=\overline{AC}\)) \(\to55^{\circ}\); \(m\angle QPR\) (when \(m\angle QRP = 30^{\circ}\) and \(\triangle PQR\) is an isosceles triangle with \(\overline{PQ}=\overline{QR}\)) \(\to30^{\circ}\); \(m\angle BDE\) (when \(m\angle BAC = 45^{\circ}\) and points \(D\) and \(E\) are the mid - points of \(\overline{AB}\) and \(\overline{BC}\) respectively in \(\triangle ABC\)) \(\to45^{\circ}\)