0
$\begingroup$

Consider two triangles $\triangle abc$ and $\triangle def$ such that $ab=de$ and $ac=df$.Also area of $\triangle abc$ is equal to area of $\triangle def$.Now draw $cm$ perpendicular to $ab$ and $fn$ perpendicular to $de$.$ab$ and $de$ are equal and area of triangles is also equal so $cm$ should be equal to $fn$.Now $\triangle amc$ and$\triangle dnf$ are congruent by right angle triangle congruence(since hypoteneous $ac$ and $df$ are equal).therefore $\angle bac$ is equal to $\angle edf$.Now in $\triangle abc$ and $\triangle def$ by SAS both $\triangle abc$ and $\triangle def$ are congruent so $bc=ef$.I don't know where i am wrong.

edit:Please read my proof and point out what's wrong

$\endgroup$
  • $\begingroup$ Look up Heron's formula. $\endgroup$ – Suzu Hirose Nov 11 '14 at 5:35
  • $\begingroup$ Your mistake is that you can’t conclude that $\angle bac=\angle edf$. You can only conclude that $\angle mac=\angle ndf$, but these equal angles are only the same angles as $\angle bac$ and $\angle edf$ if $m$ is between $a$ and $b$ and $n$ is between $d$ and $e$. This may not be the case, as you can see from Mick’s picture. The perpendicular from a triangle’s vertex to the opposite side is not always inside the triangle. $\endgroup$ – Steve Kass Nov 16 '14 at 18:22
4
$\begingroup$

The answer is no. See the following counter-example. The two triangles have the same altitude, and equal bases (and hence equal in area) but the third sides (i.e. BC, EF) are different.

This fact can also be verified by applying the formula:- area of a triangle = 0.5 ab sin C.

enter image description here

$\endgroup$
1
$\begingroup$

Consider sweeping side ab away from ac so angle bac increases from zero to $\pi$ When bac is zero, the area is zero. It increases with angle bac for a while, but when the angle gets to $\pi$ the area is back to zero again. There is some angle bac where the area is maximized. For lower area, there must (by the intermediate value theorem) be one angle bac below the one that makes maximum area and one above it.

$\endgroup$
-1
$\begingroup$

Yes I know you got confused to it. but before you go further you should know that your both triangle are acute angle and hence your altitude exist inside the triangle. but when you draw two triangle one acute and one obtuse then your altitude get outside the triangle and you are unable to prove that third side is equal. Your explanation is possible only when both triangle would be acute angled

$\endgroup$
  • $\begingroup$ Welcome to stackexchange. I'm glad you want to help people by answering questions, but you should do that for new questions that don't already have good answers. Your answer here is essentially a duplicate of both of the others. That's why it was downvoted. $\endgroup$ – Ethan Bolker Oct 29 '17 at 13:02

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.