Question

drag the tiles to the correct boxes to complete the pairs.
match the statements with their values.
$m\angle abc + m\angle bac + m\angle acb$ when $\triangle abc$ is an isosceles triangle with $\overline{ab} = \overline{ac}$
$m\angle abc$ when $m\angle bac = 70\degree$ and $\triangle abc$ is an isosceles triangle with $\overline{ab} = \overline{ac}$
$m\angle qpr$ when $m\angle qrp = 30\degree$ and $\triangle pqr$ is an isosceles triangle with $\overline{pq} = \overline{qr}$
$m\angle bde$ when $m\angle bac = 45\degree$ and points $d$ and $e$ are the midpoints of $\overline{ab}$ and $\overline{bc}$, respectively, in $\triangle abc$
$55\degree$
$180\degree$
$45\degree$
$30\degree$

Explanation:

Step1: Sum of triangle angles

The sum of interior angles of any triangle is always $180^\circ$.

Step2: Calculate $\angle ABC$ (isosceles $\triangle ABC$)

Given $\overline{AB}=\overline{AC}$, so $\angle ABC = \angle ACB$. Using $\angle BAC=70^\circ$:
$\angle ABC = \frac{180^\circ - 70^\circ}{2} = 55^\circ$

Step3: Calculate $\angle QPR$ (isosceles $\triangle PQR$)

Given $\overline{PQ}=\overline{QR}$, so $\angle QPR = \angle QRP=30^\circ$.

Step4: Calculate $\angle BDE$ (midpoints $D,E$)

$D,E$ are midpoints, so $DE \parallel AC$ by midline theorem. Corresponding angles are equal: $\angle BDE = \angle BAC=45^\circ$.

Answer:

$55^\circ$ → $m\angle ABC$ when $m\angle BAC = 70^\circ$ and $\triangle ABC$ is an isosceles triangle with $\overline{AB} = \overline{AC}$
$180^\circ$ → $m\angle ABC + m\angle BAC + m\angle ACB$ when $\triangle ABC$ is an isosceles triangle with $\overline{AB} = \overline{AC}$
$45^\circ$ → $m\angle BDE$ when $m\angle BAC = 45^\circ$ and points $D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{BC}$, respectively, in $\triangle ABC$
$30^\circ$ → $m\angle QPR$ when $m\angle QRP = 30^\circ$ and $\triangle PQR$ is an isosceles triangle with $\overline{PQ} = \overline{QR}$