In a triangle ^@ABC^@, ^@AD^@ is a median. ^@F^@ is a point on ^@AC^@ such that the line ^@BF^@ bisects ^@AD^@ at ^@E^@. If ^@AD = 9 \space cm^@ and ^@AF = 3 \space cm^@, find the measure of ^@AC^@.
Answer:
9 cm
- We are given that ^@AD^@ is the median of ^@\triangle ABC^@ and ^@E^@ is the midpoint of ^@AD.^@
Let us draw a line ^@DG^@ parallel to ^@BF^@. - Now, in ^@\triangle ADG^@, ^@E^@ is the midpoint of ^@AD^@ and ^@EF \parallel DG.^@
By converse of the midpoint theorem we have ^@F^@ as midpoint of ^@AG.^@ @^ \begin{aligned} \implies & AF = FG && \ldots \text(1) \end{aligned}@^
Similarly, in ^@\triangle BCF^@, ^@D^@ is the midpoint of ^@BC^@ and ^@DG \parallel BF.^@
By converse of midpoint theorem we have ^@G ^@ is midpoint of ^@ CF. ^@ @^ \begin{aligned} \implies FG = GC && \ldots \text(2) \end{aligned}@^ - From equations (1) and (2), we get @^\begin{aligned} AF = FG = GC && \ldots \text(3) \end{aligned}@^ Also, from the figure we see that @^ \begin{aligned} AF + FG + GC = AC \\ AF + AF + AF = AC && \text { [from (3)] } \\ 3 AF = AC \end{aligned}@^
- We are given that ^@AF^@ = 3 cm.
Thus, ^@AC = 3 AF = 3 \times 3 \text{ cm} = 9 \text{ cm}^@.
