Show that $AC=AD+BD$ in the triangle below

Question

The triangle $ABC$ in the figure below is isosceles with base $AC$, knowing that $AD$ is bisector of $\angle BAC$ and $\angle ABC=100^o$, prove that $AC=AD+BD$.

$\frac{AB}{BD}=\frac{AC}{CD}$

$\frac{BD}{DE}=\frac{AB}{AE}$

$\frac{AB}{AF}=\frac{BC}{CF}$

$AD =\sqrt{AB.AC-BD.CD}$

Answer 1

As shown in the figure above, construct $E$ on $AC$ such that $\triangle ABD\cong\triangle AED$, and $F$ on $\overrightarrow{AD}$ such that $BD=DF$. We have $$\angle ADE=\angle ADB=\angle CDE=\angle CDF=60^{\circ}\,.$$ Since $$DE=DF,\quad \angle CDE=\angle CDF,\quad CD=CD\,,$$ we can say that $\triangle CDE\cong \triangle CDF$. Hence $\angle DCF=\angle DCE=40^{\circ}$. It turns out that $$\angle ACF=\angle AFC=80^{\circ}\,,$$ implying that $\triangle ACF$ is an isosceles. Finally, $$AC=AF=AD+DF=AD+BD\,.$$

Answer 2

Draw the isosceles triangles $\triangle{AA'C}$, $\triangle{BDD'}$ and $\triangle{DA'C}$.

We have $AD=AD'$ and $BD=DD'=D'A'$. We are done.

Answer 3

As an alternative solution using the law of sines, we have

$$\frac{BD}{\sin 20°}=\frac{AD}{\sin 100°} \implies AD+BD =AD\cdot \left(1+\frac{\sin 20°}{\sin 100°}\right)$$

and

$$\frac{AC}{\sin 120°}=\frac{AD}{\sin 40°} \implies AC= AD\cdot\frac{\sin 120°}{\sin 40°}$$

then all boils down to prove that

$$\frac{\sin 120°}{\sin 40°}=1+\frac{\sin 20°}{\sin 100°}$$

which is true indeed

$$\frac{\sin 120°}{\sin 40°} =\frac{3\sin 40°-4\sin^3 40°}{\sin 40°} =3-4\sin^2 40°$$

$$1+\frac{\sin 20°}{\sin 100°}=1+\frac{2\sin 10°\cos 10°}{\cos 10°}=1+2\sin 10°$$

and since

$$\sin^2 40° =\sin^2 \left(\frac{90°-10°}{2}\right)=\frac{1-\cos{(90°-10°)} }{2}=\frac{1-\sin10° }{2}$$

we have

$$\frac{\sin 120°}{\sin 40°}=3-4\sin^2 40°=3-4\cdot \frac{1-\sin10° }{2}=1+2\sin 10°=1+\frac{\sin 20°}{\sin 100°}$$

that is $AC=AD+BD$.

Answer 4

This is rather simple and doesn't require a complicated construction or system of equations::

I believe no words are needed here, but just in case:

Locate $E$ on $AC$ and connect it with $D$ such that $AD=AE$. It is obvious that $\angle ADE=\angle AED=80^\circ$. It is also obvious that $DE=EC$. Notice that quadrilateral $AEDB$ is cyclic. Connecting $E$ and $B$ via $BE$, and using the properties of cyclic quadrilaterals, we see that $\angle EBD=\angle BED=20^\circ$. Therefore $BD=DE=EC$, therefore $AC=AD+BD$