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Consider two triangles ABC and DEF.AB=DE and AC=DF .Also area of triangle ABC is equal to the area of triangle DEF.If we draw an altitude (to one of the equal sides) in both triangles, is it(altitude) necessary to be equal in length.If not then please can anyone provide me an example.

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My actual problem is as follows. If there are two triangles with two equal sides and equal area i have somehow come to the conclusion that the third side is also equal(which is not true as the third side can be unequal).In the two triangles suppose altitude CM(to AB) and altitude FN(to DE) are drawn.Now consider triangle ACM and DFN.They have equal hypotenuse and an equal side.So by RHS congruency, triangle ACM and triangle DFN are congruent.So we can say that angle BAC and angle EDF are equal.Now consider triangle ABC and DEF, we have one angle equal and two pair of sides equal so by SAS congruency,ABC and DEF are congruent and so BC=EF.Where am i wrong?

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Hint: the area is half the product of one side of a triangle and an altitude drawn to that side.

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    $\begingroup$ You mean to say they will be equal.right? $\endgroup$ – user3819404 Nov 7 '14 at 7:25
  • $\begingroup$ @user3819404 yes. $\endgroup$ – TZakrevskiy Jun 13 '15 at 10:43
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There are always two solutions, acute and obtuse triangles that have the same two sides and same area.

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    $\begingroup$ There is a single solution when you have a right-angle. $\endgroup$ – Mark Bennet Feb 7 '15 at 7:49
  • $\begingroup$ Only if the right angel between the sides (isn't this obvious based on the basic claim?). This will happen when the area is half the multiplication of the sides, or maximum area for such two sides. $\endgroup$ – Moti Feb 7 '15 at 16:42

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