A geometry problem by Stalin Sahoo

Given the information I suspected an auxiliar construction of an isosceles triangle with sides $AD$ or $BC$ . After some exploration I found an interesting equilateral triangle and the solution came easily as follows.

The figure shows the construction. Draw a circle with center $A$ and raduis $AD$ . Extend $AC$ to $E$ . Then $\triangle ABC$ and $\triangle ADE$ are similar and $BC||DE$ .

Now let $F,G$ in the circle such that $BF\perp DE$ and $CF\perp DE$ .

Is easy to show that $\triangle AFG$ is equilateral and $BCGF$ is a rectangle.

Now $\angle AED=40$ and $\overset{\frown}{DG}=80$ . Therefore $\angle DEG=40$ . $DE$ is perpendicular bisector of $CG$ and $BF$ .

On the other hand $\overset{\frown}{DF}=20$ , then $\angle DGF=10$ . And because $\triangle DGC$ is isosceles we conclude $\angle BCD=10$ .

HiddenTriangle

We construct equilateral $\Delta AEC$ as shown in the picture. After that, we have $EA=ED=ED$ . We can easily prove $\Delta AED=\Delta ABC$ and thus $\angle AED=100^o$ . That means $\angle DEC=160^o$ . Because $ED=EC$ , $\angle DCE$ must be $10^o$ . Now we have $\angle ACB+\angle BCD+\angle DCE=\angle ACE=60^o$ , that means $\angle BCD=10^o$

Let $AB=AC=x,$ $BC=y$ and $BD=y-x.$ $\angle BCD = \theta$ [all angles are measured in degrees ] In $\Delta ABC,$ $\frac{x}{sin(40)} = \frac{y}{sin(100)} \implies x=y \frac{sin(40)}{sin(100)}$

In $\Delta BCD,$ $\frac{y-x}{sin( \theta)} = \frac{y}{sin(40- \theta)} \implies \frac{1- \frac{sin(40)}{sin(100)}}{sin(\theta)} = \frac{1}{sin(40- \theta)}$ $\implies \frac{sin(100)-sin(40)}{sin(100)}= \frac{sin(\theta)}{sin(40- \theta)}$ $\implies \frac{cos(70)}{sin(100)} =\frac{sin(\theta)}{sin(40- \theta)}$

Using componendo-dividendo, we get $\frac{sin(40- \theta)+ sin (\theta)}{sin(40- \theta)- sin (\theta)} = \frac{sin(100)+sin(20)}{sin(100)-sin(20)}$ Solving, we get $tan(20- \theta) = \frac{tan(40) tan(20)}{tan(60)} \implies \theta = \boxed{10}$

In triangle BCD, angles BCD + BDC will be equal to 40 degrees and DBC is 140 degrees so possible numerical values and combinations will be (BCD+BDC=40 degrees) combiantions are 20,20 not possible 10,30 can be a possible combination. now angle ACD= 40+10=50 angle BDC=30 degrees so in triangle ADC, angles ADC(30)+ACD(50) and angle DAC(100) sum up to 180

Solved easily!! without going into further calculations.

Construct point $E$ such that $\triangle AED \cong \triangle BAC$ and $\overleftrightarrow { AE }$ cuts $\triangle ABC$ into 2 pieces. We have $m\angle DAE = 40°$ because base angles are congruent in an isosceles triangle, so $m\angle EAC = 60°$ . Because $\overline { AE } \equiv \overline { AC }$ , $\triangle AEC$ is equilateral. Therefore, $\overline { EC } \equiv \overline { AE }\equiv \overline {ED}$ . Therefore, $\triangle DEC$ is isosceles.

$m\angle DEC=m\angle DEA+m\angle AEC=100°+60°=160°$ . Therefore, $m\angle ECD=\frac{180°-160°}{2}=10°$ . $m\angle BCD = m\angle ACE - m\angle ACB - m\angle ECD = 60°-40°-10°=10°$ .

check the link to see my solution picture link

put Ac=x,Ad=Bc=y,DC=z&angle (bcd)=t,from geometry of triangle, cos40=y^2+x2-x^2\2xy,y=2x.cos40&cos100=x^2+4x^2cos^2(40)-z^2 \2x.2x.cos40 so 4cos100.cos40=1+4cos^2(40)-4cos100.cos40-z^2\x^2. z^2=3.879385x^2,z=1.9696154447.x ,,cos(t+40)=z^2+x^2-y^2(2zx)=3.879385x^2+x^2-4cos^2(40)(2×1.9696514447x.x)=.6247877 so t+40=50 ,t=10°#####