Given that $ABC$ is a triangle, $|AD| = 9$, $|AB| = |AC| = 6$. Find the value of $|DB| \cdot |DC|$
Since $|AB| = |AC| = 6$, I thought that $|BC| = 6$. However, that truly seems to be wrong. Any helps will be appreciated.
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Can anyone take a look? – Melz Jul 19 '18 at 20:29
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$|BC|=6$ only if $ABC$ is an equilateral triangle – gd1035 Jul 19 '18 at 20:38
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@gd1035 Yes, I missed that. – Melz Jul 19 '18 at 20:41
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You might try Law of Sines or Law of Cosines. – John Jul 19 '18 at 20:42
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@John That would be tough for me. However, I can give it a try. – Melz Jul 19 '18 at 20:43
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Let $M$ be the mid-point of $BC$, let $u = |DM|$, $v = |BM| = |CM|$ and $h = |AM|$.
Since $|AB| = |AC|$, $\triangle DAM$ and $\triangle BAM$ are right angled triangles. This leads to
$$u^2 + h^2 = |DA|^2 = 9^2\quad\text{ and }\quad v^2 + h^2 = |BA|^2 = 6^2$$ As a result,
$$|DB||DC| = (u-v)(u+v) = u^2 - v^2 = (u^2+h^2) - (v^2+h^2) = 9^2 - 6^2 = 45$$
achille hui
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Let $\angle ABC=\alpha$. Then in $\Delta ABC$ and $\Delta ADB$, by the Law of Cosine, one has $$ |AC|^2=|AB|^2+|BC|^2-2|AB||BC|\cos\alpha \tag{1}$$ and $$ |AD|^2=|AB|^2+|DB|^2-2|AB||DB|\cos(180^\circ-\alpha). \tag{2}$$ From (1), one obtains $$ \cos\alpha=\frac{|BC|}{12}\tag{3}$$ and from (2), one obtains $$ \cos\alpha=\frac{45-|DB|^2}{12|DB|}.\tag{4}$$ Hence from (3) and (4), one has $$ \frac{45-|DB|^2}{12|DB|}=\frac{|BC|}{12}\tag{3} $$ or $$ |DB||BC|+|DB|^2=45 $$ which implies $$ |DC||DB|=45. $$
xpaul
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