Finding the length of the sections of this side

Question

Let there be a triangle $ABC$ with angle bisector $AL$ dividing $BC$ into $BL$ and $LC$. Can it be proven that $BL = \dfrac{ac}{b+c}$ and $LC = \dfrac{ab}{b+c}$?

Motivation: I'm trying to solve a problem from Geometry Revisited which is as follows:

Prove that $AL^2 = bc\bigg(1 - \big(\frac{a}{b+c}\big)^2 \bigg)$

My progress has been showing that $BL = \dfrac{\sin (\frac{A}{2}) }{\sin L}c$ and $LC = \dfrac{\sin (\frac{A}{2}) }{\sin L}b$ (following from the Sine rule)

Let $k = \dfrac{\sin (\frac{A}{2}) }{\sin L}$

Then using Stewart's theorem we arrive at (omitting some steps) $AL^2 = bc\bigg(1 - \big(\frac{\sin (\frac{A}{2}) }{\sin L} \big)^2 \bigg)$

Of course this is not the end result so I looked at the hint at the back and this is what it said:

Use Stewart's theorem with $m=kc$, $n=kb$, $k=a/(b+c)$.

(Here $m = BL$ and $n = LC$)

And I don't understand how such a substitution came to be.

Answer 1

We have $$AB:AC=BL:LC\tag1$$ from which we have $$BL=\frac{ac}{b+c}\quad \text{and}\quad LC=\frac{ab}{b+c}.$$

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Proof for $(1)$ :

Let $D$ be a point on $AB$ such that $CD$ is parallel to $AL$.

Then, we have $$\angle{ADC}=\angle{BAL}$$ $$\angle{ACD}=\angle{LAC}$$ Since $\angle{BAL}=\angle{LAC}$, $$\angle{ADC}=\angle{ACD}\implies AC=AD\tag2$$ Since $AL$ is paralle to $DC$, $$AB:AD=BL:LC$$ From $(2)$, $$AB:AC=BL:LC$$

Answer 2

The altitudes from $L$ of the triangles $LAB$ and $LAC$ have the same length (as $AL$ is the angular bisector of $\angle BAC$). Therefore, the area ratio $\frac{[LAB]}{[LAC]}$ is the ratio of the bases (with respect to the altitudes from $L$), which is $\frac{AB}{AC}=\frac{c}{b}$.

On the other hand, if we take $AL$ as the common base of $LAB$ and $LAC$, then let $BX$ and $CY$ be the perpendiculars from $B$ and $C$ to the line $LA$. Note that $\frac{BX}{CY}=\frac{BL}{CL}=\frac{m}{n}$ since $BLX$ and $CLY$ are similar triangles. With $ABL$ and $ACL$ having the same base $AL$, the area ratio $\frac{[LAB]}{[LAC]}$ is the ratio of the altitudes $\frac{BX}{CY}=\frac{BL}{CL}=\frac{m}{n}$.

Thus, $$\frac{c}{b}=\frac{AB}{AC}=\frac{[LAB]}{[LAC]}=\frac{BL}{CL}=\frac{m}{n}\,.$$ This shows $BL=m=\frac{ac}{b+c}$ and $CL=n=\frac{ab}{b+c}$.

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Alternatively, let $L'$ be the second intersection of the line $AL$ with the circle centered at $C$ with radius $CL$ (where we set $L':=L$ if $CL\perp AL$, namely, when this circle is tangent to $AL$). Then, $ABL$ and $ACL'$ are similar triangles. Hence, $$\frac{AB}{AC}=\frac{BL}{CL'}=\frac{BL}{CL}\,.$$

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One can prove $$AL^2+BL\cdot CL=AB\cdot AC$$ without Stewart's Theorem, which immediately leads to $$AL^2=bc\left(1-\left(\frac{a}{b+c}\right)^2\right)\,.$$ Let $T$ be the second intersection of the line $AL$ with the circumcircle of $ABC$. Then, using the power of the point $L$ with respect to the circumcircle of $ABC$ or observing that $BTL$ and $ACL$ are similar triangles yields $\frac{BL}{LT}=\frac{AL}{CL}$, or $$AL\cdot LT=BL\cdot CL\,.$$

Now, the triangles $ABT$ and $ALC$ are similar, whence $\frac{AT}{AB}=\frac{AC}{AL}$. That is, $$AL^2+AL\cdot LT=AT\cdot AL=AB\cdot AC\,.$$ Because $AL\cdot LT=BL\cdot CL$, the claim follows.

Answer 3

THE ANGLE BISECTOR THEOREM

Let the bisector of $\angle BAC$ intersect segment $\overline{BC}$ at point $L$. Extend ray $\overrightarrow{AL}$ to point $D$ such that line $\overleftrightarrow{CD}$ Is parallel to line $\overleftrightarrow{AB}$.

Then line $\overleftrightarrow{AD}$ is a transversal of the parallel lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$. Hence $\alpha = m\angle BAD = m\angle CDA$. Clearly $m\angle BLA = m\angle CLD$. It follows that $\triangle BLA \sim \triangle CLD$. Hence $\dfrac{AB}{BL} = \dfrac{DC}{CL}$; that is $\dfrac cm = \dfrac fn$.

Because line $\overleftrightarrow{AL}$ is an angle bisector, $m\angle BAL = m\angle CAL = \alpha$. It follows that $m\angle BAL = m\angle CDL = \alpha$. Hence $\triangle ACD$ is an isoceles triangle. So $b = AC = CD = f$.

We conclude that $\dfrac cm = \dfrac bn$.

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Because of the angle bisector theorem, we know that $\dfrac bc= \dfrac nm$. If we let $n = bk$, then we must also have $m = ck$. So we get $a = m + n = bk + ck = (b+c)k$ and, so, $k = \dfrac{a}{b+c}$. Hence $m = \dfrac{ac}{a+b}$ and $n = \dfrac{ab}{b+c}$.

For the above triangle, Stewart's theorem states that

$$b^2m + c^2n = a(d^2 + mn)$$

Thus \begin{align} b^2m + c^2n &= a(d^2 + mn) \\ \dfrac{ab^2c}{b+c} + \dfrac{abc^2}{b+c} &= a\left(d^2 + \dfrac{a^2bc}{(b+c)^2}\right) \\ abc &= a\left(d^2 + \dfrac{a^2bc}{(b+c)^2}\right) \\ d^2 &= bc\left(1 - \dfrac{a^2}{(b+c)^2} \right)\\ \end{align}