The following question was there in a euclidean text:
- Construct triangle ABC, if [ABC]/(2s-a) = 3/2 and tA = 5 and b = 6. Note that [S] its area of figure S and tX refers to the length of the bisector of angle of vertex X.
I attempted to solve the question by method of finding the locus and begun by constructing the auxiliary figure. I was able to create side b and the angle bisector but am still wondering how the third given expression could help. I will appreciate any help to solve this question with proper steps.
The key point is to find radius of inscribed circle $r$, we have:
$\frac S{2p-a}=\frac 32 \Rightarrow S=3p-\frac32 a$
$r=\frac S p=3-\frac{3a}{2p}=3\big(\frac {2p-a}{2p}\big)=3\big(\frac{b+c}{2p}\big)$
Now we try few numbers for $r$, among them only $r=2$ fits the condition, we have:
$3(\frac{b+c}{2p})=2\Rightarrow \frac{b+c}{2p}=\frac 2 3 $
$4p=3(b+c)\Rightarrow 6p-3(b+c)=2p\Rightarrow 3[2p-(b+c)=a]=2p $
which gives:
$a=\frac{2p}3$
$S=3p-\frac 32 a=2p$
which gives:
$r=\frac S p= \frac{2p}p=2$
which confirms our guess. Now to construct the triangle , consider the fact that the foot of bisector of angle BAC is on the point the circle with radius $5$ is tangent with inscribed circle and edge BC. So, as can be seen in figure, follow this:
1- Draw AB=b=6.
2-Draw a circle center at A , with radius 5.
3- Draw a line from C tangent on this circle, the tangent point is G.
4- Draw a circle with radius 2 tangent with AB and previous circle. For this, connect A to G and draw a line parallel with AC at distance 2. The intersection of these two lines is the center I of circle. , the tangent point G is the foot of bisector of $\widehat {A}$.
5- Draw a line from A tangent on inscribed circle (r=2), the intersection of this line with line from C is vertex B.
I checked the measures in my figure:
$a=BC=91.98, h=AH=49.72, c=AB=76.7, b=60$, we have:
$S=2226.8$ and $p=114.34$
$r=\frac {2286.8}{114.34}=20$
Note the scale of figure is 10/1.