The Converse of the Pythagoras Theorem

Question

Let $ABC$ be a triangle in the plane. Suppose that $AB^2+AC^2=BC^2.$ Prove that:

$\angle BAC$ is right angle.

Remark: I believe this to be true. Now I have the following difficulty. I'm looking for a proof which a 10th grade student can understand, but any effort of mine is via proof by contradiction. I mean assuming that the given angle is not a right angle and then arrive on a contradiction. So I'm looking for a solution which could avoid the indirect method but if not then a solution by the indirect method.

We habe $AB^2 + AC^2 = BC^2 + 2AB\cdot BC \cdot \cos(\angle ABC)$ — martini, Aug 16 '13 at 10:59
What about Euclid's proof? It's Proposition 48 in first book, you can find the whole book for free here: http://farside.ph.utexas.edu/euclid.html — pppqqq, Aug 16 '13 at 10:59
This can be quickly and easily be proved using the law of cosines, yet a 10th grader perhaps hasn't studied that much trigonometry... — DonAntonio, Aug 16 '13 at 10:59
http://europa.sim.ucm.es/compludoc/AA?a=Casey%2c+Stephen&donde=otras&zfr=0 — JP McCarthy, Aug 16 '13 at 11:13
What is that link, @JpMcCarthy ? Either it malfunctions or else...well, it looks weird. — DonAntonio, Aug 16 '13 at 11:16
@DonAntonio It is a proof of the converse without using Pythagoras Theorem itself. My classmate in college did it and the link is to the Mathematical Gazette citation. — JP McCarthy, Aug 16 '13 at 11:22
@DonAntonio Casey, Stephen; Mathematical Gazette; 2008; 92 (524); 309-312; 00255572. — JP McCarthy, Aug 16 '13 at 11:23
Still don't get it, @JpMcCarthy : how can that paper be reached?! — DonAntonio, Aug 16 '13 at 14:30
@DonAntonio I think you'll need a hard copy. I am not being as helpful as I had intended! — JP McCarthy, Aug 16 '13 at 14:31
Oh, so the paper doesn't exist in PDF form in that site (or any other)? — DonAntonio, Aug 16 '13 at 14:31
@DonAntonio He doesn't have it anymore. I will go into the Maths department in UCC and scan it (in a month or two --- I haven't forgotten you!) — JP McCarthy, Oct 04 '13 at 10:55
@DonAntonio The plot thickens. I went into the Maths department, into the room where I definitely saw the issue of the Mathematical Gazette which housed the proof... that one issue is missing! I will email one of the professors who I know helped submit the proof: I am not sure how often he checks his email so I will write to him in two weeks if he hasn't replied. Regards. — JP McCarthy, Oct 26 '13 at 08:30
Finally!! This is the best I can do: http://jpmccarthymaths.files.wordpress.com/2013/10/pythagoras-converse0001.pdf — JP McCarthy, Oct 30 '13 at 20:22
@DonAntonio Steve was an excellent student: top of our class. Sadly he was lost to physics! — JP McCarthy, Oct 30 '13 at 20:23

score 6 · Accepted Answer · answered Aug 16 '13 at 11:05

Draw side by side the given triangle $\;\Delta ABC\;$ and a new triange $\,PQR\;$ as follows:

$$PQ=AB\;,\;\;PR=AC\;,\;\;\angle QPR=90^\circ$$

Now, we're given $\,AB^2+AC^2=BC^2\;$, and Pythagoras theorem also gives $\,PQ^2+PR^2=QR^2\;$ , so

$$BC^2=AB^2+AC^2=PQ^2+PR^2=QR^2\implies BC=QR$$

and we get $\,\Delta ABC\cong\Delta PQR\;$ by $\;S.S.S.\;$

score 3 · Answer 2 · answered Aug 16 '13 at 11:00

3

You can use cosine formula as

$BC^2$ = $AB^2 + AC^2 -2.(AB).(AC)cos(A)$

substituting the condition we get $2.(AB).(AC)cos(\angle BAC)=0$

which gives us $\angle$ $BAC =\pi/2$

answered Aug 16 '13 at 11:00

Shobhit

6,902

score 0 · Answer 3 · answered Dec 29 '16 at 21:42

0

a^2+b^2= c^2 in triangle ABC, and argue towards contradiction as Draw the altitude of the triangle towards one of the sides making angle C, and call the length from vertix C to this intersection x. Then x^2+h^2= b^2 and (a-x)^2+h^2= c^2. Playing around with these three equations shows you that x= 0 which is a contradiction

answered Dec 29 '16 at 21:42

Tesfa

1

1

Welcome to Math.SE! You may want to see here for how to correctly format your posts. Also, could you elaborate on your proof to make it complete and easy to follow? – shardulc Dec 29 '16 at 21:50

score 0 · Answer 4 · answered Dec 29 '16 at 23:31

Well, what always worked for me as a tenth grader was to cut a triangle into right triangles:

Construct a point $D$ on $BC$ so that $DA$ is perpendicular to $BC$ so $ADC$ is a right triangle and $AD^2 + DC^2 = AC^2$.

THere are three possibilities.

i) $\angle ABC$ is acute.

Then $D$ is between $B$ and $C$ and $BC = BD + DC$.

Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; BC^2 = BD^2 + 2BD*DC + DC^2$

And $AB^2 + BC^2 = AD^2 + DC^2 + 2(BD^2+ BD*DC) = AC^2 + 2(BD^2+ BD*DC) > AC^2$

(It helps if you draw a picture and note $AB > AD; BC > BD$ so of course $AB^2 + BC^2 > AD^2 + DC^2 = AC$. But the calculation tells you exactly how much bigger.)

ii) $\angle ABC$ is obtuse.

Then $B$ is between $D$ and $C$ and $DC = DB + BC$.

Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; DC^2 = DB^2 + 2DB*BC + BC^2$

And $AB^2 + BC^2 = AD^2 + DC^2 - 2(BD^2+ BD*DC) = AC^2 - 2(BD^2+ BD*DC) < AC^2$

iii) $\angle ABC$ is right.

Then $B=D$ and $C$ and $DC = BC $.

Then $AB^2 = AD^2; AC^2 = AD^2 + DC^2=AB^2 + BC^2; BC^2 = DC^2 $

So i) $AB^2 + BC^2 > AC^2 \implies \triangle ABC$ is acute and $AB^2 + BC^2 = AC + 2(BD^2 + BD*DC)$.

ii) $AB^2 + BC^2 < AC^2 \implies \triangle ABC$ is abtuse and $AB^2 + BC^2 = AC - 2(BD^2 + BD*DC)$.

iii) $AB^2 + BC^2 = AC^2 \implies \triangle ABC$ is right.

.... which gives the tenth grader a hint of the law of cosines to come.

[i.e. $\pm(BD^2 + DB*DC) = AB*BC*\cos \angle ABC$ which is $0$ if $\angle ABC$ is right; positive if obtuse and negative if acute.]

score -1 · Answer 5 · answered Dec 01 '14 at 07:47

Understand Converse of Pythagoras Theorem with a VIDEO explanation. Click on the link to WATCH the VIDEO: WATCH VIDEO

Converse of Pythagoras Theorem

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Given : A triangle ABC such that AB square plus BC square equals to AC square. To prove : the triangle ABC is right - angled at B. Construction: Construct a right - angled triangle PQR, right - angled at Q such that PQ equal to AB and QR equals to BC. Prof: Appling the Pythagoras theorem to right - angled triangle PQR we have PQ square plus QR square equals to PR square or AB square plus BC square is equal to PR square since PQ is equal to AB and QR equals to BC by construction. Let this be equation 1 but it is given that AB square plus BC square equals to AC square. From equations 1 and 2 we get PR square equals to AC square i.e., PR equals to AC. Therefore, by SSS congruence criterion, we get triangle ABC is congruent to triangle PQR so we get angle B equal to angle Q but we know angle Q is equal to 90 degrees by construction. Therefore, angle B equals to 90 degrees. Hence, triangle ABC is a right triangle, right - angled at B. Thus, the theorem is proved.

The Converse of the Pythagoras Theorem

5 Answers5