In an isosceles triangle, the median joining the vertex (formed by intersection of equal sides) to the midpoint of the opposite side is an altitude.

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Detailed Solution

in the triangle ΔABC :
AB=AC & ∠B=∠C since ΔABC is isosceles.
BD=CD , since AD is the median and a median divides the side it is projected on in half
Now taking ΔABD and ΔADC
AB=AC
BD=CD
∠B=∠C
Therefore ΔABD and ΔADC are congruent triangles
⇒ΔABD ≅ ΔADC (By Side Angle Side Theorem(SAS))
we can conclude that,
⇒∠ADB=∠ADC (Common parts of Congruent Triangles(CPCT)) .....(i)
BDC is a straight line therefore
∠BDC=180∘
⇒∠ADB+∠ADC = ∠BDC = 180∘
Substituting equation (i) in the above equation, we get
⇒2∠ADB=180∘
⇒∠ADB=90∘
⇒∠ADB=∠ADC=90∘
An altitude is a line that makes an angle of ∠90∘ with the line it is projected on, therefore it is proved that ADAD is the altitude of the isosceles triangle ΔABC

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