Construct a triangle given certain lengths related to a bisector

Question

Let $ABC$ be a triangle, and $AD$ the bisector of angle $A$. Write $AB = c$, $AC = b$, $AD = d$, $BD = c'$, $CD = b'$. Using ruler and compass, construct the triangle $ABC$ given the lengths $d$, $b-b'$ and $c - c'$. (That is, we are given segments of lengths $d$, $b-b'$ and $c-c'$ to work with, and the problem is to reconstruct the triangle from these data.)

I already have an analytic solution to the problem that can be translated into a lengthy construction. I am looking for a geometric solution which is as simple as possible.

Answer 1

I assume that $\angle BAC$ is also given. Let $AX = x = b – b’$ and $AY = y = c – c’$. By angle bisector theorem, x : y = b’ : c’.

(1). Construct the circle $\alpha$ (centered at A, radius = d).

(2) Locate $B_1$ (exterior to ⊿ABC) on $\alpha$ such that $\angle B_1AB = \angle BAD$. $C_1$ is similarly constructed.

(3) B and C will be somewhere on the perpendicular bisectors of $B_1D$ and $C_1D$ respectively.

(4) B and C will be somewhere on the perpendicular bisectors of $B_1Y$ and $C_1X$ respectively.

(3) and (4) determine length and location of the line BC.

Answer 2

1) line $AX=x,AY=y,AN=d ,J $ is midpoint of $XY, M $is midpoint of $AX$,

2)circle $J @r=JY,AK \perp AX$, cross circle$J$ at $K$

3)make rectangle$AKQN$, connect $AQ, KL \perp AQ$ , cross $AQ$ at $L$

4) connect $ML,QB$//$ML$ cross $AN$ at $B$

5)circle $B@r=XB$, circle $A@r=AN$, two circle cross at $D$

6)circle $D@r=DN$, cross circle $A@r=AN$ at $P$

7)connect $BD,AP$, cross at $C$

$\triangle ABC$ is the wanted. note $AN> AK$ is necessary condition.

the proof is to OP.

edit: the op comments remind me that there is more simple way to build $AB$

$AG=\dfrac{d^2}{x}+y$

Answer 3

Let $ABC$ be a triangle, and $AD$ the bisector of angle $A$. Write $|AB|=c$, $|AC|=b$, $|AD|=d$, $|BD|=c'$, $|CD|=b'$. Using ruler and compass, construct the triangle $ABC$ given the lengths $d$, $b-b'=u$ and $c-c'=v$. (That is, we are given segments of lengths $d$, $b-b'=u$ and $c-c'=v$ to work with, and the problem is to reconstruct the triangle from these data.)

A line $\mathcal{L}_1$ through an arbitrary point $D$.

Points $\{A,D_2\}=\mathcal{C}_1(D,d)\cap\mathcal{L}_1$.

Point $D_3=\mathcal{C}_2(D_2,d)\cap\mathcal{L}_1$.

Points $\{p_1,p_2\}=\mathcal{C}_3(D_3,|DD_3|)\cap\mathcal{C}_4(D,|DD_3|)$.

A line $\mathcal{L}_2$ through points $p_1,p_2$.

Point $p_3=\mathcal{C}_4(D_2,v)\cap\mathcal{L}_2$.

Points $\{p_4,p_5\}=\mathcal{C}_5(p_3,|p_3p_2|)\cap\mathcal{C}_6(D,|p_3p_2|)$.

A line $\mathcal{L}_3$ through points $p_4,p_5$.

Point $p_6=\mathcal{L}_2\cap\mathcal{L}_3$.

Point $p_7=\mathcal{C}_7(p_6.|p_3p_6|)\cap\mathcal{L}_2$.

Point $\{p_8,p_9\}=\mathcal{C}_8(p_7,u)\cap\mathcal{L}_2$.

Points $\{p_{10},p_{11}\}=\mathcal{C}_9(D_2,|D_2p_8|)\cap\mathcal{C}_{10}(A,|D_2p_9|)$.

A line $\mathcal{L}_4$ through points $A,p_{10}$.

A line $\mathcal{L}_5$ through points $A,p_{11}$.

Points $\{p_{12},p_{13}\}=\mathcal{C}_{10}(A,|D_2p_9|)\cap\mathcal{C}_{11}(p_{10},|D_2p_9|)$.

A line $\mathcal{L}_6$ through points $p_{12},p_{13}$.

Point $C=\mathcal{L}_6\cap\mathcal{L}_4$.

A line $\mathcal{L}_7$ through points $C,D$.

Point $B=\mathcal{L}_5\cap\mathcal{L}_7$.

A line $\mathcal{L}_1$ through an arbitrary point $D$.

Points $\{A,D_2\}=\mathcal{C}_1(D,d)\cap\mathcal{L}_1$.

Point $D_3=\mathcal{C}_2(D_2,d)\cap\mathcal{L}_1$.

Points $\{p_1,p_2\}=\mathcal{C}_3(D_3,|DD_3|)\cap\mathcal{C}_4(D,|DD_3|)$.

A line $\mathcal{L}_2$ through points $p_1,p_2$.

Point $p_3=\mathcal{C}_4(D_2,v)\cap\mathcal{L}_2$.

Points $\{p_4,p_5\}=\mathcal{C}_5(p_3,|p_3p_2|)\cap\mathcal{C}_6(D,|p_3p_2|)$.

A line $\mathcal{L}_3$ through points $p_4,p_5$.

Point $p_6=\mathcal{L}_2\cap\mathcal{L}_3$.

Point $p_7=\mathcal{C}_7(p_6.|p_3p_6|)\cap\mathcal{L}_2$.

Point $\{p_8,p_9\}=\mathcal{C}_8(p_7,u)\cap\mathcal{L}_2$.

Points $\{p_{10},p_{11}\}=\mathcal{C}_9(D_2,|D_2p_8|)\cap\mathcal{C}_{10}(A,|D_2p_9|)$.

A line $\mathcal{L}_4$ through points $A,p_{10}$.

A line $\mathcal{L}_5$ through points $A,p_{11}$.

Points $\{p_{12},p_{13}\}=\mathcal{C}_{10}(A,|D_2p_9|)\cap\mathcal{C}_{11}(p_{10},|D_2p_9|)$.

A line $\mathcal{L}_6$ through points $p_{12},p_{13}$.

Point $C=\mathcal{L}_6\cap\mathcal{L}_4$.

A line $\mathcal{L}_7$ through points $C,D$.

Point $B=\mathcal{L}_5\cap\mathcal{L}_7$.

The sought triangle is $\triangle ABC$.

EDIT

Basic idea is to construct $\triangle ADC$ on given base $AD=d$ using the analytic solution $|AC|=b=\tfrac12\left( \frac{d^2}{v}+u \right) $ and $|DC|=b'=b-u$.

It seems to be easier to construct $\triangle AD_2p_{10}$ with doubled sides first and then construct point $C$ as a middle of $Ap_{10}$.

The point $p_{10}$ is constructed as intersection of circles with sides $|D_2p_9|=2b=\frac{d^2}{v}+u$ and $|D_2p_8|=2b'=\frac{d^2}{v}-u$.

Since $AD$ is the bisector of $\angle CAB$, the line through $A,B$ (or through $A,p_{11}$) is symmetric to the line through $A,C$ with respect to $AD$, thus the point $B$ is found at the intersection of lines through $C.D$ and $A,p_{11}$.

The key question is then to construct a length $|D_2p_7|=\frac{d^2}v$ by means of a circle through three points, $D,D_3$ and $p_3$, where we have $d^2=|DD_2|\cdot|D_2D_3|=|D_2p_3|\cdot|D_2p_7|=v\cdot \frac{d^2}v$. The center of the circle is the intersection of the midpoint perpendicular to $Dp_3$ with the line through $p_1,p_2$, and point $p_7$ is obviousely found at the intersection of the circle and line through $p_1,p_2$.

Answer 4

a more nice method to rebuild the triangle:

$AD=d,AK=c-c',AJ=b-b', F $ is mid point of $AD$

1) circle $F@r=AF$,circle $A@r=AK$,circle $A@r=AJ$, three circles cross at $K_1,J_1$

2)$PQ \perp AD, AK_1$ cross $PQ$ at $P$,cross circle $AJ$ at $J_2, AJ_1$ cross $PQ$ at $Q$, cross circle $AK$ at $K_2$

3) take midpoint $M$ of $QK_2$, midpoint $N$ of $PJ_2$, connect $MN$ ,cross $AD$ at $I$

4)circle $I@r=ID$, cross circle $AK$ at $K_3$,cross circle$AJ$ at $J_3$

5)circle $A@r=AM,A@r=AN$, connect $AK_3$ cross circle $AM$ at $B$, connect $AJ_3$ , cross circle $AN$ at $C$.

$\triangle ABC$ (or $\triangle AB_1C_1$) is the one wanted.