Relation between AD, BD and BC

Question

In a triangle $ABC$, $\sphericalangle BAC = 100°$ , $AB=AC$. A point $D$ is chosen on the side $AC$ such that $\sphericalangle ABD = \sphericalangle CBD$, prove that $AD+DB=BC$.

I tried solving this problem making an imaginary figure in my mind. I found all the angles according to given conditions.

$\sphericalangle ABD= \sphericalangle CBD=20°$, $\sphericalangle ACB=40°$, $\sphericalangle ADB=60°$, $\sphericalangle BDC=120°$

Using the angle bisector theorem, we get the relation between sides as $AB/BC=AD/CD$ but I could not get the idea to simplify further.

Answer 1

Inspired by YNK's figure, I find a new approach as follows:

In the figure, $E$ is a point of $BC$ such that $BE=BD$.

By isosceles triangle, $\angle BED=80^{\text o}$.

Hence $\angle EDC=\angle ECD=40^{\text o}$. Consequently

$$ED=EC \tag{1}$$. On the other hand, $\angle BAD+\angle BED=180^{\text o} $

$\therefore BADE$ is a cyclic quadrilateral.

Since $\angle ABD=\angle EBD=20^{\text o} $ and $BADE$ is a cyclic quadrilateral, $$ED = AD \tag{2}$$

$(1)$ and $(2) \implies$

$$EC=AD \tag{3}$$

Hence $$BC=BE+EC=BD+AD$$

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Note: We can prove that $ED=AD$ by $2$ different methods.

Method $(1)$: Since $ABED$ is a cyclic quadrilateral, we can draw a circle passing through the $4$ points.

Let $O$ be the centre.

$$\angle EOD=\angle DOA=40^{\text o} \implies ED=AD$$

Method $(2)$, by using Sine Rule

in $\Delta EBD$, $$\frac{ED}{\sin 20^{\text o}}=\frac{BD}{\sin 80^{\text o}} \tag{3}$$

in $\Delta ABD$, $$\frac{AD}{\sin 20^{\text o}}=\frac{BD}{\sin 100^{\text o}} \tag{4}$$

Note that $\sin 80^{\text o}=\sin 100^{\text o}$

Hence $$(3), (4) \implies AD=ED.$$

Answer 2

Hint (Synthetic way): Let $E,F\in BC $ such that $\angle BDE=60^{\circ}$ and $\angle CDF=40^{\circ}$.

Your way: Using the angle bisector theorem for the triangle $\triangle ABC$ and the angle bisector $BD,$ we have $\frac{DA}{AB}=\frac{DC}{CB}$. This gives $\frac x{x+y}=\frac{y}{BC}$ and $$BC=\frac{y}x(x+y)\tag1$$ where $x=DA$, $y=DC$.

By law of cosines on the triangles $\triangle BAD$ and $\triangle BCD$, we have $$DB^2+x^2-DB\, x=(x+y)^2\tag2$$ and $$DB^2+y^2+DB\, y=BC^2.\tag3$$ If we substract $(2)$ from $(3)$, we get $$DB=\frac{BC^2}{x+y}-2y. \tag4$$ If we use $(1)$ in $(4)$ we get $$DB+AD=DB+x=\frac{x^3+y^3-2x^2y+xy^2}{x^2}.\tag5$$ On the other hand, drawing the altitude $AE$ of the isoceles triangle $\triangle BAC$ of the base side $BC$ and using $(1)$, we have $$\cos 40^{\circ}=\cos(\angle ABC)=\frac{EC}{AC}=\frac{BC/2}{(x+y)}=\frac{y}{2x}.$$ Now, $z=\cos 40^{\circ}$ satisfies the equation $8z^3-6z+1=0$. (See the note below.) Hence $$8\left(\frac{y}{2x}\right)^3-6\left(\frac{y}{2x}\right)+1=0\implies \bbox[yellow,5px,border:2px solid red]{x^3+y^3=3x^2y}.$$ Putting this in $(5)$ gives $$DB+AD=\frac{3x^2y-2x^2y+xy^2}{x^2}=\frac yx(x+y)=BC.$$ Note: Let $z=\cos40^\circ$. Using the triple angle formula $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\cos(3\times40^\circ)=\cos120^\circ=-\frac12$, we have $4z^3-3z=-\frac12$ or $8z^3-6z+1=0.$

Answer 3

To facilitate the sought proof, we need to enhance the scenario described in OP’s problem statement as shown in the diagram. We have marked point $E$ on the side $BC$ such that $BE=BD$. We have introduced a circular arc with radius equal to $BC$, the center of which is at the vertex $B$, to cut extended $BA$ and $BD$ at $G$ and $H$. Finally, after joining $H$ to $C$ and $G$, a line is drawn through $D$ parallel to $HG$ to meat $BG$ at $F$.

The proof follows as soon as the missing angles marked with small arcs are found through angle chasing. We would like you to work out the proof yourselves using the hints we have given here. In case you find it still too difficult to do it, let us know and we will oblige with the complete solution.

$\underline{\text{Edit to Add the Proof Upon OP’s Request}}:$

$\angle GAC$ and $\angle CAB$ are supplementary angles, so we have, $$\measuredangle GAC = 180^o - \measuredangle CAB =180^o - 100^0=80^o. \tag{1}$$

Since $BH$ and $BG$ are radii of the added arc, $\triangle GBH$ is an isosceles triangle. We have drawn $DF$ parallel to $HG$. Hence, $\triangle FBD$ is also an isosceles triangle and that makes $\measuredangle DFB=80^o$. we have already shown that $\measuredangle CAB=80^o$ (see (1)). As a consequence, $\triangle ADF$ is an isosceles triangle. Therefore, we have, $$AD = FD. \tag{2}$$

Applying the intercept theorem to $\triangle GBH$ and the segment $FD$, we shall write, $$\dfrac{FD}{GH}=\dfrac{BD}{BH}. \tag{3}$$

Because they are radii of the arc, the two sides of $\triangle HBC$, i.e., $BC$ and $BH$, are equal. Since we have selected $E$ on the side $BC$ such that $BE=BD$, segment $ED$ is parallel to $CH$. Therefore, we can apply the intercept theorem to $\triangle HBC$ and the segment $FD$ to obtain, $$\dfrac{ED}{CH}=\dfrac{BD}{BH}. \tag{4}$$

The two chords $CH$ and $GH$ of the arc subtend equal angles at the center $B$ of the arc, $${\large{\pmb{\therefore}}}\quad CH=GH. \tag{5}$$

From (2), (3), (4), and (5), we can deduce that, $$ED=FD=AD. \tag{6}$$

$\triangle BED$ is another isosceles triangle, because $BD=BE$. Therefore, we know that, $$\measuredangle BED=\dfrac{180^o-\measuredangle DBE}{2}=80^o.$$

Since $\angle BED$ is an external angle of $\triangle DEC$, we have, $$\measuredangle CDE = \measuredangle BED - \measuredangle ECD =80^o-40^0=40^o.$$

Since $\measuredangle ECD$ of $\triangle DEC$ is also $40^o$, this triangle is an isosceles triangle as well. $${\large{\pmb{\therefore}}}\quad CE=DE. \tag{7}$$

Using (6) and (7), we can show that $CE=AD$. Finally, we have, $$AD+DB=CE+EB=CB.$$

Answer 4

A nice way to "reverse engineer" your problem (and the idea comes form the internal angles of $\triangle ABC$) is to include the original picture in the frame of a regular nonagon, as shown above.

Since $BD$ bisects $\angle ABC$ and $\triangle BTC$ is isosceles, your thesis is reached once we show that $AD \cong TD$. To this aim note that

1. By angle chasing we find that $\triangle AVB$ is isosceles, hence $AB \cong VB$.
2. Angle chasing and AAA criterion yields $\triangle VBC \sim\triangle TDC$, which gives $$\frac{\overline{TD}}{\overline{DC}} = \frac{\overline{VB}}{\overline{BC}}.$$
3. Internal Bisector Theorem on $\triangle ABC$ yields $$\frac{\overline{AD}}{\overline{DC}} = \frac{\overline{AB}}{\overline{BC}}.$$
4. Using 1. and comparing 2. and 3. gives $AD \cong TD$ as desired.

$\blacksquare$