Answer:
See explanation
Step-by-step explanation:
Q1. 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}[/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
Q2. Statement Reason
1. [tex]\overline {LM}\perp \overline {MO},\\ \\\overline {PN}\perp \overline {MO}[/tex] Given
2. [tex]\angle ONP, \ \angle OML\text{ are right angles}[/tex] Definition of perpendicular
3. [tex]\angle ONP\cong \angle OML[/tex] All right angles are congruent
4. [tex]\angle O\cong \angle O[/tex] Reflexive property
5. [tex]\triangle LMO\cong \triangle PNO[/tex] AA similarity theorem
Q3. Statement Reason
1. [tex]ABCD\text{ is a trapezoid}[/tex] 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 DAC\cong \angle ACB[/tex] Alternate interior angles are congruent (Alternate interior angles theorem)
5. [tex]\triangle AED\sim \triangle CEB[/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{BA}}[/tex] Transitive property
6. [tex]\triangle ABE\sim \triangle DBC[/tex] SAS similarity theorem
