# Calculate length of one side given all other sides in a triangle.

All the lengths in red had been given in the problem statement. The task is to find $BC$. The green ones were calculated from the information in red.

Here's two ways I tried to solve it (and I would like feedback):

1. The green line is drawn parallel to $QC$. Suppose, the other unlabelled end of the green line be point $R$. So the line can be called $PR$.

In $\triangle BPR$,

$PC=QC=9, \space\space PC=3\times AP\\PB=6, \space\space PB=3 \times AQ\\\therefore \triangle BPR \sim \triangle APQ\\BD=4\times3, \space\space \boxed{BD=12}$

In quadrilateral $PQCD$, opposite sides are parallel to each other and it is therefore a parallelogram. So $DC=4$.

$BC=BD+DC\\BC=12+4\\\boxed{BC=16}$

1. In $\triangle ABC$, $AB=2+6=2(1+3)=\boxed{2\times 4}\\AC=3+9=3(1+3)=\boxed{3\times 4}\\BC=?$

In $\triangle APQ$, $AP=\boxed{2}\\AC=\boxed{3}\\PQ=4$

$\because \triangle ABC\sim \triangle APQ,\\BC=PQ\times4=4\times4=16,\\\boxed{BC=16}$

• The red $9$ above the green line should be written in green, right? – 5xum Jan 24 '18 at 9:41
• Also, when you wrote $PC=QC$, I think you meant $PR=QR$. – 5xum Jan 24 '18 at 9:42
• @5xum Problem fixed. Thanks for noticing! – Soha Farhin Pine Jan 24 '18 at 9:45
• I still see $PC=QC$ which is not true. – 5xum Jan 24 '18 at 9:48
• I don't see the point D anywhere – Darkrai Jan 24 '18 at 9:53

There are so many typos that it's kind of hard to read your proof.

1. You write $PC=QC$, but you probably mean $PR=QR$.
2. You write $PC=3\cdot AP$, which is not true.
3. You write $PB = 3\cdot AQ$ which is not true (since $6\neq 3\cdot 3$)
4. You write $\Delta BPR\sim\Delta APQ$ which is almost ok, but not entirely - be careful to write the corners in the correct order!
5. In the middle of your proof, you start talking about a point called "$D$" you did not previously define (i.e. "in the quadrilateral $PQCD$", there is no previous mention of $D$ - you probably meant $PQCR$.

Your question doesn't mention $PQ$ is parallel to $BC$ but your answer does.
So assuming that line $PQ$ is parallel to $BC$ then by similar triangles
( Considering the triangles $APQ$ and $ABC$) $$\frac {AP}{AB}= \frac {AQ}{AC}=\frac {PQ}{BC}$$ Hence we get $$\frac {3}{12}=\frac {4}{BC}$$ $$\Rightarrow BC=16$$