Question

given: $overline{ps}congoverline{qs},overline{dp}congoverline{lq}$ prove: $\triangle dsl$ is isosceles. 1. given 2. base angles theorem 3. $angle1,angle2$ and $angle3,angle4$ are linear pairs. 4. $angle1,angle2$ and $angle3,angle4$ are supplementary. 4. if two angles are a linear pair, then they

Explanation:

Step1: State given information

$\overline{PS}\cong\overline{QS},\overline{DP}\cong\overline{LQ}$

Step2: Apply Base - Angles Theorem

Since $\overline{PS}\cong\overline{QS}$, then $\angle SPQ=\angle SQP$

Step3: Define linear pairs

$\angle 1,\angle 2$ and $\angle 3,\angle 4$ are linear pairs

Step4: Use property of linear pairs

If two angles are a linear pair, they are supplementary, so $\angle 1 = 180^{\circ}-\angle 2$ and $\angle 4 = 180^{\circ}-\angle 3$

Step5: Substitute equal angles

Since $\angle SPQ=\angle SQP$, substitute to get $180^{\circ}-\angle 1=180^{\circ}-\angle 4$

Step6: Simplify using subtraction property

Subtract 180° from both sides to get $\angle 1=\angle 4$

Step7: Prove triangle is isosceles

Since $\angle 1=\angle 4$ in $\triangle DSL$, $\triangle DSL$ is isosceles

Answer:

1. $\overline{PS}\cong\overline{QS},\overline{DP}\cong\overline{LQ}$
2. $\angle SPQ=\angle SQP$ (Base - Angles Theorem since $\overline{PS}\cong\overline{QS}$)
3. Definition of linear - pairs
4. Definition of supplementary angles
5. $\angle 1 = 180^{\circ}-\angle 2$ and $\angle 4 = 180^{\circ}-\angle 3$
6. Since $\angle SPQ=\angle SQP$, then $180^{\circ}-\angle 1=180^{\circ}-\angle 4$ (Substitution)
7. $\angle 1=\angle 4$ (Subtraction Property of Equality)
8. $\triangle DSL$ is isosceles (If two angles of a triangle are congruent, then the triangle is isosceles)