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Given : $\triangle ABC$, $AB=7$, $AC=9$, the angle bisector of $\angle BAC$ passes through $BC$ in point $D$ such that $AD = BD$, find $BC$. Here is the drawing: enter image description here

I have no idea where to start with this one.

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$$\frac{BD}{DC}=\frac{AB}{AC}\Rightarrow BD=7x=AD, CD=9x$$ $$AD^2=AC\cdot AB-BD \cdot CD$$ $$49x^2=63-63x^2$$ $$112x^2=63$$ $$x^2=\frac{9}{16}$$ $$x=\frac34$$ $$BC=16x=16\cdot \frac 34=12$$

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  • $\begingroup$ By a tedious computation, I got $\tan\alpha=\sqrt5/2$, and you got $\cos\alpha=2/3$, so I have to agree. But I don’t see the reason for your first ratio. Could you remind me? $\endgroup$
    – Lubin
    Apr 22, 2016 at 20:00
  • $\begingroup$ en.wikipedia.org/wiki/Angle_bisector_theorem $\endgroup$
    – Roman83
    Apr 22, 2016 at 20:05
  • $\begingroup$ Oh yes, now I remember seeing this relatively late in life; but Mrs. Stewart didn’t teach us that one. Thanks. $\endgroup$
    – Lubin
    Apr 22, 2016 at 20:52

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