Point $D$ in $\triangle{ABC}$ such that $\angle{DBC}=\angle{DCB}=10^{\circ}$, $\angle{BAD}=20^{\circ}, \angle{DAC}=40^{\circ}$, show that $AD=AC$

Question

Point $D$ In $\triangle{ABC}$ such that $\angle{DBC}=\angle{DCB}=10^{\circ}, \angle{BAD}=20^{\circ}, \angle{DAC}=40^{\circ}$, show that $AD=AC$. Wonder if there is pure geometric approach to prove this statement? Thanks

For trigonometric solution, apply Ceva's theorem in angles: Let $\angle{ABD}=x$

$$ \begin{multline}\nonumber \shoveleft \dfrac{\sin20}{\sin40} \dfrac{\sin(100-x)}{\sin10} \dfrac{\sin10}{\sin x}=1 \\ \shoveleft \implies \dfrac{\sin(100-x)}{2\sin x \cdot \cos20}=1 \\ \shoveleft \implies \cos(x-10)=2\sin x \cdot \cos20\\ \shoveleft \implies \cos x \cdot \cos10+\sin x \cdot \sin10=2\cos20 \cdot \sin x\\ \shoveleft \implies \tan x = \dfrac{\cos10}{2\cos20-\sin10}=\dfrac{\sin30 \cdot \cos10}{\cos20 -\sin30 \cdot \sin10}\\ \shoveleft =\dfrac{\sin30 \cdot \cos10}{\cos20-\tfrac{1}{2}(\cos20 + \cos40)}=\dfrac{\sin30 \cdot \cos10}{\tfrac{1}{2}(\cos20-\cos40)}\\ \shoveleft = \dfrac{\sin30 \cdot \cos10}{\cos30 \cdot \cos10}=\tan30 \implies x=30^{\circ} \implies \angle{ACD}=70^{\circ} \implies AC=AD \blacksquare \end{multline} $$

Answer 1

Let $E$ be a point on the circumcircle of $ABD$ such that $DB=DE$. Then $\angle DBE = \angle BAD = 20^\circ$. Note that $B, C, E$ lie on a circle with center $D$. Hence $\angle CDE = 2\angle CBE = 2\cdot 30^\circ = 60^\circ$. It follows that triangle $CDE$ is equilateral. In particular, $DE=CE$. Since $\angle DAE = 20^\circ$, we also have $\angle EAC = 20^\circ$. This shows that $E$ lies both on the internal bisector of $\angle DAC$ and on the perpendicular bisector of $CD$. This can only happen if $AD=AC$ or if $AD\neq AC$ and $E$ is the midpoint of the arc $CD$ of the circumcircle of $ADC$ (but this is not the case since $E$ and $A$ lie on the same side of the line $CD$). Therefore $AD=AC$.

Answer 2

We can "reverse" the problem by showing first that $ABC$ can be embedded in a regular nonagon as shown above.

Conditions on $\measuredangle BAD$ and $\measuredangle DAC$ are obviously satisfied.

As for $\measuredangle DBC$ and $\measuredangle DCB$, observe that $PQD$ is equilateral by construction. Hence $QCD$ is isosceles, and angle chasing yields $\measuredangle DCQ = 50^\circ$. Hence $\measuredangle DBC = 10^\circ$. The same can be done at the other side, of course.

Thus the construction corresponds to the one in OP.

Now it is an easy matter of angle chasing to determine $\measuredangle DCA = 70^\circ$, from which the thesis.

$\blacksquare$

Answer 3

Another approach: We draw a circle on diameter AB. The reflects of AB about AD and also BD about BC meet at E which is on the circle, because:

$\angle EAC=40-20=20=\angle DBE$

Suppose the extion of BD meets the circle at F. Angle FDC is the exterior angle of triangle DBC so $\angle FDC=20^o$

in this way :

$\angle FDC=\angle FBE=20^o\Rightarrow DC||BE$

$AE\bot BE \Rightarrow AE\bot DC$

That is AE is bisector of angle DAC and also the height of triangle ADC which means triangle ADC is isosceles and $AC=AD$

Note that $\angle CAE=20=FAE$, so F is also on AC.This means AC and extension of BD meet at F which is on the circle.