1
$\begingroup$

Given triangle as given in the below image, assume that angles $BCA, CA'B$ and $AB'C$ have equal measure. Show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$. What additional assumption gives the Pythagorean Theorem?sally Roots to research pg 62 2.10 ii, Howard Eves- History of math., pg 215, 7.8(b)


Need first show that if angles $BCA, CA'B$ and $AB'C$ have equal measure, show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$.
Consider the $3$ triangles: $BCA, CA'B, AB'C$, with orientation shown in the same cck order, and the common angle in center, with base being respectively $AB, BC, CA$.
It is obvious that only similarity can be worked out here.


Edit : Kindly see my first comment below for doubts about whether only similarity can be worked out here.


*Edit 2: * Using the @GCab hint: triangles $BCA, CA'B$ common angles: $C=A', B=B$ (but, now the 2nd triangle would be listed as $BA'C$ to have the same cck orientation of angles and sides). Unable to find any (need one similar side between common angles, to apply similarity) common sides. Anyway the third angle of both triangles have to be equal to each other. Similarly, for triangles $BCA, CB'A$. Can use two ratios for these two pairs of slr. triangles (i.e., (i)$BCA, BA'C$. , (ii) $BCA, CB'A$) to get : $\frac{AB}{CB} = \frac{CB}{A'B}$, $\frac{AB}{AC} = \frac{AC}{AB'}$. This would lead to two equalities among products to be summed up, to solve the first part easily.

But, what is the intuition gained?
The intuition I expected was in algebraic terms, and not that for second question to be solved $A'=B'$, i.e. there can be only one position on $AB$ that can raise at $90^0$ to meet vertex $A$, even if I am correct.

$\endgroup$
  • 1
    $\begingroup$ The additional assumption is, obviiously, that $\angle BCA=90^\circ$ (which makes $A'=B'$) $\endgroup$ – Hagen von Eitzen Mar 31 '18 at 13:56
  • $\begingroup$ @HagenvonEitzen But, this would bring more understanding if the first question is completely answered. I have left it complete. Please help in that. Also, am I correct in stating that only similarity can be worked out, it should be case (triangle's dimensions', angles' based) dependent. But, not sure. $\endgroup$ – jiten Mar 31 '18 at 13:57
  • 1
    $\begingroup$ sum of angles $=\pi$ for all the triangles $\endgroup$ – G Cab Mar 31 '18 at 14:01
  • $\begingroup$ @GCab Using hint: triangles $BCA, CA'B$ common angles: $C=A', B=B$ (but, now the 2nd triangle would be listed as $BA'C$ to have the same cck orientation of angles and sides). Unable to find any (need one similar side betn. cmn. angles, to apply similarity) common sides. Anyway the third angle of both triangles have to be equal to each other. Similarly, for triangles $BCA, CB'A$. Can use two ratios for these two pairs of slr. triangles (i.e., (i)$BCA, BA'C$. , (ii) $BCA, CB'A$) to get : $\frac{AB}{CB} = \frac{CB}{A'B}$, $\frac{AB}{AC} = \frac{AC}{AB'}$. But, what is the intuition gained? $\endgroup$ – jiten Mar 31 '18 at 14:27
  • $\begingroup$ @GCab Please help with the intuition gained with this question, as the issue is raised in detail in the second Edit of the OP. $\endgroup$ – jiten Mar 31 '18 at 14:41
1
$\begingroup$

Tr_quasi_Pit_1

The triangles $ACB$, $AB'C$ and $BA'C$ are similar, having two angles (and one edge) in common.
Thus establish the similarity proportionalities $$ \left\{ \begin{gathered} \frac{{c_{\,a} }} {b} = \frac{d} {a} = \frac{b} {c} \hfill \\ \frac{{c_{\,b} }} {a} = \frac{d} {b} = \frac{a} {c} \hfill \\ \end{gathered} \right. $$ which then can be developed as follows $$ \eqalign{ & \left\{ \matrix{ ac_{\,a} = bd \hfill \cr bc_{\,b} = ad \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cac_{\,a} = bab \hfill \cr cbc_{\,b} = aab \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cc_{\,a} = b^{\,2} \hfill \cr cc_{\,b} = a^{\,2} \hfill \cr dc = ab \hfill \cr} \right. \cr & c\left( {c_{\,a} + c_{\,b} } \right) = a^{\,2} + b^{\,2} \cr} $$ to obtain the starting relation in your post.

Then to obtain that $c_a+c_b=c$ you shall clearly have that $\gamma = \pi/2$.

$\endgroup$
  • $\begingroup$ Please state which software, if open-source, is used to draw image. $\endgroup$ – jiten Apr 1 '18 at 9:32
  • $\begingroup$ Also, I hope the only mathematical intuition is that of the law of cosines, i.e. $c^2=a^2+b^2-2ab\cos C$, with angle $C$ changing from acute to obtuse, with being a right angle in between. Also, this applying for the special case of $3$ common angles, as shown above. However, the algebraic significance of the similarity of $3$ common angles is unknown still. May be application of ratios applies here, which couples with the law of cosines. $\endgroup$ – jiten Apr 1 '18 at 9:50
  • $\begingroup$ @jiten: a) the drawing has been realized with Geogebra, which is open-source. b) Sure, you can rework the problem in many related ways, including cosine law, but also sine law etc. c) "..significance of the similarity of $3$ common angles is unknown still" : don't catch what you mean. $\endgroup$ – G Cab Apr 1 '18 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.