Answer:
See explanation
Step-by-step explanation:
Q1. Statement Reason
1. ABCD is a trapezoid Given
2. [tex]\overline{AD}\parallel \overline{BC}[/tex] Definition of trapezoid
3. [tex]\angle ADE\cong \angle CBD[/tex] Alternate interior angles are congruent (Alternate interior angles theorem)
4. [tex]\angle DAE\cong \angle BCA[/tex] Alternate interior angles are congruent (Alternate interior angles theorem)
5. [tex]\triangle AED\sim \triangle CEB[/tex] AA similarity theorem
Q2. Statement Reason
1. [tex]T\text{ is the midpoint of }\overline {QR}\\ \\U\text{ is the midpoint of }\overline {QS}\\ \\V\text{ is the midpoint of }\overline {RS}[/tex] Given
2. [tex]\overline {TV},\ \overline {TU},\ \overline {UV},\\ \\\text{are midsegments of }\triangle QRS[/tex] Midsegments connect midpoints of opposite sides
3. [tex]\overline{TU}=\dfrac{1}{2}\overline{RS}\\ \\\overline{UV}=\dfrac{1}{2}\overline{QR}\\ \\\overline{VT}=\dfrac{1}{2}\overline{SQ}[/tex] Triangle midsegment theorem
4. [tex]\dfrac{\overline{TU}}{\overline{RS}}=\dfrac{1}{2}\\ \\\dfrac{\overline{UV}}{\overline{QR}}=\dfrac{1}{2}\\ \\\dfrac{\overline{VT}}{\overline{SQ}}=\dfrac{1}{2}[/tex] Division property of equality
5. [tex]\dfrac{\overline{TU}}{\overline{RS}}=\dfrac{\overline{UV}}{\overline{QR}}=\dfrac{\overline{VT}}{\overline{SQ}}[/tex] Transitive property
6. [tex]\triangle QRS\sim \triangle VUT[/tex] SSS similarity theorem
Q3. Statement Reason
1. [tex]\overline{LM}\perp\overline{MO},\ \overline{PN}\perp\overline{MO}[/tex] Given
2. [tex]\angle OMP,\ \angle OML\\ \\\text{ are right angles}[/tex] Definition of perpendicular
3. [tex]\angle OMP\cong \angle OML[/tex] All right angles are congruent
4. [tex]\angle O\cong \angle O[/tex] Reflexive property
5. [tex]\triangle LMO\sim \triangle PNO[/tex] AA similarity theorem
Q4. Statement Reason
1. [tex]\overline {AC}\text{ and }\overline{EC}\\ \\\text{intersect at }B[/tex] Given
2. [tex]\angle ABE\cong \angle CBD[/tex] Vertical angles theorem
3. [tex]\dfrac{\overline{BC}}{\overline{EB}}=\dfrac{5}{15}=\dfrac{1}{3}[/tex] Proportion of corresponding sides
4. [tex]\dfrac{\overline{BD}}{\overline{AB}}=\dfrac{4}{12}=\dfrac{1}{3}[/tex] Proportion of corresponding sides
5. [tex]\dfrac{\overline{BC}}{\overline{EB}}=\dfrac{\overline{BD}}{\overline{AB}}[/tex] Transitive property
6. [tex]\triangle ABE\sim \triangle DBC[/tex] SAS similarity theorem
