In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$. If $AP = 3$,$PB = 5$ and $AQ = 6$,find $QC$ and $AC$.
(N/A) In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.
By the Basic Proportionality Theorem (Thales Theorem):
$\frac{AP}{PB} = \frac{AQ}{QC}$
Given $AP = 3$,$PB = 5$,$AQ = 6$.
$\frac{3}{5} = \frac{6}{QC}$
$QC = \frac{6 \times 5}{3} = 10$.
Now,to find $AC$:
$AC = AQ + QC = 6 + 10 = 16$.
Alternatively,using the corollary $\frac{AP}{AB} = \frac{AQ}{AC}$:
$\frac{3}{3+5} = \frac{6}{AC}$
$\frac{3}{8} = \frac{6}{AC}$
$AC = \frac{6 \times 8}{3} = 16$.
By the Basic Proportionality Theorem (Thales Theorem):
$\frac{AP}{PB} = \frac{AQ}{QC}$
Given $AP = 3$,$PB = 5$,$AQ = 6$.
$\frac{3}{5} = \frac{6}{QC}$
$QC = \frac{6 \times 5}{3} = 10$.
Now,to find $AC$:
$AC = AQ + QC = 6 + 10 = 16$.
Alternatively,using the corollary $\frac{AP}{AB} = \frac{AQ}{AC}$:
$\frac{3}{3+5} = \frac{6}{AC}$
$\frac{3}{8} = \frac{6}{AC}$
$AC = \frac{6 \times 8}{3} = 16$.
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