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Draw ΔABC and let D be the midpoint of AB¯. Let E be the midpoint of CD¯. Let F be the intersection of AE→ and BC¯. DrawDG¯ parallel toEF¯ meetingBC¯ at G. Prove that BG=GF=FC .

Short Answer

Expert verified

The diagram is:

The midpoint divides the segment into two congruent segments and Transitive Property proves BG=GF=FC.

Step by step solution

01

Step 1. Consider the diagram.

The ΔABCis:

In ΔABC, D be the midpoint of AB¯. Let E be the midpoint of CD¯. Let F be the intersection of AE→and BC¯. Draw DG¯parallel to EF¯ meetingBC¯ at G.

02

Step 2. State the transitive property.

The transitive property states that if A=B and B=C, then A=C .

03

Step 3. Show the proof.

Statement

Reasons

In ΔDGC, E is the midpoint of CD¯and DG¯∥EF¯.

Given.

F is the midpoint of GC¯.

A line that contains the midpoint of the side of a triangle and is parallel to another side passes through the midpoint of the third side.

GF=FC

Midpoint divides the segment into two congruent segments.

In ΔABF, D is the midpoint of AB¯and DG¯∥AF¯

Given.

G is the midpoint of BF¯.

A line that contains the midpoint of the side of a triangle and is parallel to another side passes through the midpoint of the third side.

BG=GF

Midpoint divides the segment into two congruent segments.

BG=GF=FC

Transitive Property.

04

Step 4. State the conclusion.

Therefore, the statement BG=GF=FCis proved by:

The midpoint divides the segment into two congruent segments that isBG=GF=FC

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