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In right triangle $ABC,$ $\angle C = 90^\circ.$ Let $P$ and $Q$ be points on $\overline{AC}$ so that $AP = PQ = QC.$ If $QB = 67$ and $PB = 76,$ find $AB.$

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How do I use ratios and given side lengths to create a proportion to solve for $AB$? Is there any other way to solve this?

I would think the best way to approach this is to relate $QB/CB = AB/CB$, though that would make $CB$ for both the same. I guess the relation of $AB/AC = QB/QC$ can also be used.

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  • $\begingroup$ The triangles aren’t similar. $\endgroup$ – Tavish May 28 at 18:54
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Let $BA= y, BC=x. $ We can solve numerically using Pythagoras thm twice.

$$ 2 (67.^2 - x^2)^{0.5} = (76^2 - x^2)^{0.5}$$

$$ 3 (67.^2 - x^2)^{0.5} = (y^2 - x^2)^{0.5}$$

Since there are 2 equations and two unknowns, we can solve by squaring and eliminating $x$.

Begin by squaring both equations. Solve for $x$ and substitute in the second set to find $y$:

$$ y=89, x=63.7181$$

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Well, we can use Pythagoras twice (as @Narasimham pointed out):

$$ \begin{cases} \text{BP}^2=\text{CP}^2+\text{BC}^2\\ \\ \text{BQ}^2=\text{CQ}^2+\text{BC}^2\\ \\ \text{CP}=\text{CQ}+\text{PQ}\\ \\ \text{CQ}=\text{PQ}\color{\white}{+}\\ \\ \text{BP}=76\\ \\ \text{BQ}=67 \end{cases}\tag1 $$

In order to solve this, I used Mathematica (it is also easy to solve by hand) with the following code:

In[1]:=BP = 76;
BQ = 67; FullSimplify[
 Solve[{BP^2 == CP^2 + BC^2, BQ^2 == CQ^2 + BC^2, CP == CQ + PQ, 
   CQ == PQ, CP > 0 && BC > 0 && CQ > 0 && PQ > 0}, {CP, BC, CQ, PQ}]]

Out[1]={{CP -> 2 Sqrt[429], BC -> 2 Sqrt[1015], CQ -> Sqrt[429], 
  PQ -> Sqrt[429]}}

So, we know:

$$\text{AB}=\sqrt{\left(2\sqrt{1015}\right)^2+\left(3\cdot\sqrt{429}\right)^2}=89\tag2$$

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Let $|BC|=a$, $|AB|=c$, $|AC|=b$, $|AP|=|PQ|=|QC|=\tfrac b3$, $|BP|=u=76$, $|BQ|=v=67$.

Applying Stewart’s Theorem, we have

\begin{align} \triangle AQB:\quad |AB|^2\,|PQ|+|BQ|^2\,|AP|-|AQ|\,(|BP^2|+|AP|\,|PQ|) &=0 ,\\ c^2\cdot\tfrac b3+v^2\cdot\tfrac b3 -2\cdot\tfrac b3\cdot(u^2+\cdot(\tfrac b3)^2) &=0 ,\\ 9c^2-2b^2-18u^2+9v^2 &=0 \tag{1}\label{1} , \end{align}

\begin{align} \triangle PCB:\quad |BP|^2\,|QC|+|BC|^2\,|PQ|-|PC|\,(|BQ^2|+|PQ|\,|QC|) &=0 ,\\ u^2\cdot\tfrac b3+a^2\cdot\tfrac b3 -2\cdot\tfrac b3\cdot(v^2+\cdot(\tfrac b3)^2) &=0 ,\\ -9a^2+2b^2-9u^2+18v^2 &=0 ,\\ -9c^2+11b^2-9u^2+18v^2 &=0 \tag{2}\label{2} . \end{align}

From \eqref{1},\eqref{2} we have

\begin{align} b^2 &= 3(u^2-v^2)=3861 , \\ c^2 &= \tfrac13\,( 8u^2-5v^2 ) =7921 . \end{align}

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Let AP = PQ = QC = $x$. Note that PB and QB are the medians of the triangles AQB and PCB respectively. Per the Apollonius's Theorem for medians

$$AB^2 + QB^2 = 2x^2 + 2PB^2\tag1 $$ $$CB^2 +PB^2 = 2x^2 + 2QB^2\tag2 $$

Take (1) - (2) and recognize $AB^2 -CB^2=(3x)^2$ to get

$$ x^2=\frac13(PB^2-QB^2)$$

Then, substitute $x^2$ in (1) to obtain

$$AB= \sqrt{ \frac83 PB^2 - \frac53 QB^2}= 89 $$

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Hint:

Suppose $AP=PQ=QC=x$. Then, we know $$BC^2 = 67^2-x^2 =76^2 -(2x)^2 $$ You can solve for $x$ from here and get $AB=3x$.

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For easy understanding, assume $AP=PQ=QC=x$ & $BC=y$ then using Pythagoras theorem in respective right triangles, we get $$QB^2=QC^2+BC^2\iff 67^2=x^2+y^2\tag 1$$ $$PB^2=PC^2+BC^2\iff 76^2=4x^2+y^2\tag 2$$ $$AB^2=AC^2+BC^2\iff AB^2=9x^2+y^2\tag 3$$ Subtracting (1) from (2) $$3x^2=76^2-67^2\iff x^2=429$$ Subtracting (1) from (3), $$AB^2-67^2=8x^2\iff AB^2=8x^2+67^2=8(429)+67^2=7921$$$$ \color{blue}{AB=}\sqrt{7921}=\color{blue}{89}$$

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