# Two triangles with two equal sides and equal area will have the third size also equal?

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.

• Look up Heron's formula. Nov 11, 2014 at 5:35
• 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. Nov 16, 2014 at 18:22 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.