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I've the following construction, as shown in the figure. The line $QP$ is tangent to the incircle of $\triangle ABC$. The triangle has side lengths given by $a,b,c$. I am trying to prove the result that $$(sp-bc)(sq-bc)=bc(s-b)(s-c)$$ where $s$ is the semiperimeter of the triangle, $p = AP$, and $q = AQ$.

Show that $(sp-bc)(sq-bc)=bc(s-b)(s-c)

I've tried applying the Law of Cosines separately for $\triangle APQ$ and $\triangle ABC$, as they share the angle $\angle CAB$, but I've had no luck.

Blue
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  • to me the LHS looks a lot smaller than the RHS. Where is this equality coming from? – Vasili Nov 09 '18 at 03:42
  • I've verified the identity with some ugly trigonometric arithmetic. I'm looking for a cleaner argument. – Blue Nov 09 '18 at 04:47

2 Answers2

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Let the usual $x=s-a=AD=AE$. We know $PQ=PE+QD=2x-p-q$, so the cosine rule gives $$ \frac{(b+c)^2-a^2}{2bc}=1+\cos A=\frac{(p+q)^2-(2x-p-q)^2}{2pq}=\frac{2x(p+q-x)}{pq}. $$ Clearing denominators, we have $$ pq((b+c)^2-a^2)=2bc\cdot 2x(p+q-x) $$ i.e. $$ pqs-bc(p+q)=-bcx. $$ Multiply by $s$ and add $b^2c^2$: $$ pqs^2-bc(sp+sq)+b^2c^2=bc(bc-xs). $$ The LHS is $(sp-bc)(sq-bc)$. In the RHS, recall $x=s-a=b+c-s$, so $$ bc(bc-xs)=bc[bc-(b+c-s)s]=bc(s-b)(s-c) $$ as desired.

user10354138
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  • Why do you have $(b+c)^2$ instead of $b^2+ c^2$? –  Nov 09 '18 at 11:51
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    To make the presentation a bit nicer. I could have done it without the +1 but then I will need to add them later in order to get rid of the $2pq$ in $p^2+q^2-(2x-p-q)^2$. Of course this suggests we shouldn't really approach the question with the cosine rule. The method by @Blue is nicer, I just did it this way to show it is possible to complete where you left off. – user10354138 Nov 10 '18 at 11:04
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Consider the following:

enter image description here

Due to pairs of congruent triangles, we can write

$$|\triangle ABC| = |\triangle APQ| + 2\,|\triangle BIP| + 2\,|\triangle CIQ| \tag{1}$$ where $I$ is the incenter of $\triangle ABC$. Thus, $$\begin{align} |\triangle ABC| - |\triangle APQ| &= r \left(\;(c - p) + (b - q)\;\right) &\left(\;|\triangle BIP|=\frac12r(c-p), \text{etc}\;\right) \tag{2}\\[6pt] s\left(\;|\triangle ABC| - |\triangle APQ|\;\right) &= |\triangle ABC|\, (b+c-p-q) &\left(\;rs=|\triangle ABC|\;\right)\tag{3}\\[6pt] s\left(\;\frac12b c \sin A - \frac12 p q \sin A\;\right) &= \frac12 b c \sin A\, (b+c-p-q) \tag{4}\\[6pt] s\left(\;b c - p q \;\right) &= b c\, (b+c-p-q) \tag{5}\\[6pt] \end{align}$$ The reader readily verifies that $(5)$ is, for $s \neq 0$, equivalent to the desired relation. $\square$

Blue
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