Question 1

&#10;Supply missing reasons for the following proof.&#10;Given:   bisects   ;   and   Prove:   S1.   bisects   ; R1.   and   S2.   R2.&#10;S3.   and   are rt.   R3.&#10;S4.   R4.&#10;S5.   R5.&#10;S6.   R6.&#10;S7.   R7.

Accepted Answer

R1. Given&#10;R2. Definition of angle-bisector&#10;R3. Perpendicular lines form right angles.&#10;R4. All right angles are congruent.&#10;R5. Identity&#10;R6. AAS&#10;R7. CPCTC

Question 2

&#10;Use the drawing provided to explain the following.&#10;Given:   ,   , and   are the medians of   Prove:

Accepted Answer

If

If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

,

, and

are the medians of If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

, then their point of concurrence is
Q, the centroid of If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

. It follows that If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

. By the Segment-Addition
Postulate, we have If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

. By substitution, If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

. By subtraction, If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

. In turn, If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

or 2. Multiplying the equation If , , and are the medians of , then their point of concurrence is Q, the centroid of . It follows that . By the Segment-Addition Postulate, we have . By substitution, . By subtraction, . In turn, or 2. Multiplying the equation by , we have .

by

, we have

.

Question 3

&#10;Supply missing statements and missing reasons for the following proof.&#10;Given: Point X not on   so that   Prove: X lies on the perpendicular-bisector of   S1. R1.&#10;S2.   R2.&#10;S3. With M the midpoint of   , R3.&#10;draw   S4. Then   R4.&#10;S5. Also,   R5.&#10;S6. R6. SSS&#10;S7.   R7.&#10;S8. R8. Definition of perpendicular-bisector of a line segment

Accepted Answer

S1. Point X not on   so that   R1. Given&#10;R2. Definition of congruent line segments&#10;R3. Through 2 points, there is exactly one line.&#10;R4. Definition of midpoint of line segment&#10;R5. Identity&#10;S6.   R7. CPCTC&#10;S8. X lies on the perpendicular-bisector of

Question 4

&#10;Use the drawing provided to explain the following theorem.&#10;&#34;The three angle bisectors of the angles of a triangle are concurrent.&#34;&#10;Given:   ,   , and   are the angle bisectors of the angles of   Prove:   ,   , and   are concurrent at point Z

Accepted Answer

The answer of  &#10;Use the drawing provided to explain...

Deck 7: Locus and Concurrence