Prove that area of $ABC$ is $\frac{1}{4}|BC|^2\cdot \cot \frac{\theta}{2}$ where $\theta=\angle BAD$

Question

Let $ABC$ be a triangle such that $\angle B=2\angle C$. We extend the side $BC$ by a segment $CD$ equal to $\frac{1}{3}BC$.

Prove that area of $ABC$ is $$\frac{1}{4}|BC|^2\cdot \cot \frac{\theta}{2}$$ where $\theta=\angle BAD$

My attempt,

Let $a,b,c,d$ be the respective lengths of $BC, CA, AB, AD$ and suppose that the bisector of angle $ABC$ intersects $AC$ at $E$. So we can see that $BE=EC=\frac{ab}{(a+c)}$. We also know that $AC:AB=BC:BE$. Hence, $b^2=c(a+c)$. By applting the Cosine Law to the triangles $ABC$ and $ACD$, we can get $ad^2+(\frac{a}{3})c^2=(b^2+\frac{a^2}{3})(\frac{4a}{3})$. By using the equation of $b^2$, we get $9d=4a^2+12b^2-3c^2=(3c+2a)^2 $and $d=\frac{1}{3}(3c+2a)$. Applying the Cosine Law again to triangle $ABD$ and using the previous result to get $$16a^2=9(c^2+d^2-2cd\cos \theta)=(18c^2+12ac+4a^2)-6c(3c+2a)\cos \theta$$ so that $$6(3c^2+2ac)\cos \theta=6(3c^2+2ac-2a^2)$$

Hence, $$\cos \theta=1-\frac{2a^2}{3c^2+2ac}$$

We can get $1+\cos \theta=\frac{6c^2+4ac-2a^2}{3c^3+2ac}$ and $\cot^2 \frac{\theta}{2}=\frac{1+\cos \theta}{1-\cos \theta}=\frac{(c+a)(3c-a)}{a^2}$

By using Heron's area formula,

$16[ABC]^2=a^4 \cot^2 \frac{\theta}{2}$

$[ABC]=\frac{1}{4}a^2 \cot \frac{\theta}{2}$

Is there any other way/method to solve this question? Thanks in advance.

Answer 1

In the picture we have as a model the triangle $\Delta ABC$ with sides $a,b,c$ equal to respectively $15,18,12$. OP also mentions the formula for $b^2$, here is an "other way" to obtain it.

After construction, geogebra confirms optically that the angle bisector in $B$, the perpendicular bisector of $BC$ and the side $AC$ are concurrent, this means that $\hat B=2\hat C$. (I will no longer use the hats, this should be clear in the given context.)

Lemma: $$\bbox[yellow]{\qquad AD = c+\frac{2a}3\ .\qquad}$$ Proof: Let $R$ be the circumradius, then: $$ \begin{aligned} b&=2R\sin B=2R\sin 2C=2R\cdot 2\sin C\cos C=2c\cos C =2c\cdot\frac{a^2+b^2-c^2}{2ab}\ ,\text{ so} \\ ab^2 &=c(a^2+b^2-c^2)\ ,\\ b^2(a-c)&=c(a^2-c^2)=c(a+c)(a-c)\ ,\text{ thus giving a formula for $b^2$,}\\ b^2&=c(a+c)\ .\\[3mm] &\qquad\text{ From here we compute $AD^2$ in $\Delta ABD$:}\\[3mm] \cos B&=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2-ac}{2ac}=\frac{a-c}{2c}\ ,\\ AD^2 &=BA^2+BD^2-2BA\cdot BD\cdot \cos B \\ &=c^2+\frac {16}9a^2-2 c\cdot\frac {4a}3\cdot\frac{a-c}{2c} =c^2+\frac 43ac+\frac 49a^2 =\bbox[yellow]{\left(c+\frac{2a}3\right)}^2\ . \end{aligned} $$ $\square$

Note: The values $15,18,12$ in the picture are so that $b^2=18^2=12(15+12)=c(a+c)$. The converse is with the same argument true, if we have the relation $b^2=c(a+c)$, then $B=2C$.

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We can start now the final computation. Recall a formula in "some other triangle" $A'B'C'$ with usual notations enlarged with the prime decorator. The sides are for instance $a',b',c'$, the semiperimeter is $s'=\frac 12(a'+b'+c')$ and so on. Then as for instance mentioned here, we have: $$ \cot\frac {A'}2 =\sqrt{\frac{s'(s'-a')}{(s'-b')(s'-c')}} =\frac{\sqrt{s'(s'-a')(s'-b')(s'-c')}}{(s'-b')(s'-c')} \overset{\text{Heron}}= \frac{\operatorname{Area}[A'B'C']}{(s'-b')(s'-c')}\ . $$ We apply this formula in the triangle $\Delta A'B'C'=\Delta ABD$, its angle in $A'$ is with the OP notations $\theta$, so we have step by step: $$ \begin{aligned} a'&=\frac{4a}3\ ,\qquad b'=\bbox[yellow]{c+\frac{2a}3}\ ,\qquad c'=c\ ,\\ s'&=\frac 12(a'+b'+c')=a+c\ , \qquad s'-b'=\frac a3\ ,\qquad s'-c'=a\ ,\\[4mm] \operatorname{Area}[ABC] &=\frac 34\operatorname{Area}[ABD] =\frac 34(s'-b')(s'-c')\cot\frac \theta 2 =\frac 34\cdot\frac a3\cdot a\cdot\cot\frac \theta 2 \\ &=\frac 14a^2\cot\frac \theta 2\ . \end{aligned} $$ $\square$

Answer 2

Since $\angle ABC \cong 2\angle ACB$ we can "carve out" triangle $ABC$ from an isosceles trapezoid $ABCE$, with $AB \cong AE \cong CE$, as in the Figure below, where, in addition, $M$ is the midpoint of $AE$, $N = AD \cap CM$, and $H$ is the projection of $A$ on the trapezoid base $BC$. Observe that $$[ABC] = [BMC] = \frac14 \overline{BC}^2 \cot \left(\frac{\measuredangle BMC}2\right),$$so that the assertion is proved once we show that $$\angle BAD \cong \angle BMC\tag{1}\label{1}.$$ We will prove that $$ABN \sim BMC,$$ which in particular implies \eqref{1}.

For simplicity let $\overline{BC} = 1$ and $\overline{AB} = x$. Applying twice Pythagorean Theorem we can first determine the altitude of the trapezoid $$\overline{AH} = \frac{\sqrt{3x^2+2x-1}}2,$$ and then $$\overline{AD} = x+\frac23.\tag{2}\label{2}$$Incidentally, observe that \eqref{2} is the expression mentioned in dan_fulea's answer, and that this expression shows $A$ lies on a hyperbola with focii in $B$ and $D$.

Similarity $AMN \sim CDN$ (AAA criterion), with similarity ratio $\frac{3x}2$, leads to $\overline{AN} = x$ and $\overline{DN} = \frac23$. As a result we have, by SAS criterion, $BND \sim CND$, yielding $\angle BND \cong \angle NCD$, and, by supplementarity, $$\angle ANB \cong\angle BCM.\tag{3}\label{3}$$Having already proved that $ABN$ is isosceles, \eqref{3} is enough to conclude that $ABN \sim BMC$ (AAA criterion), as desired.

$\blacksquare$