<s>[INST] How prove two segments are congruent If $\angle ACB=2\angle ABC=4\angle BAD$,and $D$ is on $BC$, such $CG\parallel AB$, $CG=CD$, $HG \parallel AC$.
show  that
$$BD=HG$$

My idea: let $\angle BAD=x$, then $\angle ABC=2x$, $\angle ACB=4x$.

Lemma 1：in $\Delta ABC,AB=c,AC=b,BC=a$  if  $\angle A=2\angle B$,then  $a^2=b(b+c)$

So use this lemma:
in $\Delta ABC$, we have
$$AB^2=AC(AC+BC)$$
and in $\Delta ABD$, we have
$$AD^2=BD(BD+AB)$$
Then I can't prove $$BD=HG.$$
Can someone help me?
 [/INST] The statement can be proved by applying the sine theorem four times.
If $\widehat{BAD}=\theta,\widehat{ABC}=2\theta,\widehat{ACB}=4\theta$, then $$\widehat{DAC}=\pi-7\theta,\widehat{DCG}=2\theta,\widehat{ADC}=3\theta,\widehat{DHG}=7\theta,\widehat{CDG}=\widehat{CGD}=\frac{\pi}{2}-\theta,\\ \widehat{HDG}=\frac{\pi}{2}-2\theta.$$
Now we consider the triangles $ABD,ADC,CDG,GDH$, in order to have:
$$ BD = \frac{\sin\theta}{\sin 2\theta}AD = \frac{\sin\theta \sin 4\theta}{\sin 2\theta \sin 7\theta}CD = \frac{\sin\theta \sin 4\theta \cos \theta}{\sin 2\theta \sin 7\theta \sin 2\theta}DG = \frac{\sin\theta \sin 4\theta \cos \theta \sin 7\theta}{\sin 2\theta \sin 7\theta \sin 2\theta \cos2\theta}HG,$$
that implies $BD=HG$ since $\sin 4\theta = 2\sin 2\theta\cos 2\theta$ and $\sin 2\theta = 2 \sin\theta \cos\theta$.
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