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Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CD

AD is an altitude, therefore angle ADB = angle ADC = 90 degrees


Then, in triangles ABD and ACD,

AD is common,

angle ADB = angle ADC

and BD = CD

Therefore the two triangles are congruent (SAS).


And therefore AB = AC, that is, the triangle is isosceles.


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Q: If the median to a side of a triangle is also an altitude to that side then the triangle is isosceles How do you write this Proof?
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