Let △ABC and △DEF are two isosceles triangles with AB = AC and DE = DF and ∠A = ∠D
Now, let AM and DN are their respective altitudes or heights.
Let △ABC and △DEF
AB/DE = AC/DF
∠A = ∠D [given]
∴ △ABC ~ △DEF [by SAS similarity]
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠B = ∠E and ∠C = ∠F ……(i)
In △ABM and △DEN
∠B = ∠E [from (i)]
and ∠M = ∠N [each 90°]
∴ △ABC ~ △DEF [by AA similarity]
So, AM/DN = AB/DE = BM/EN ……(ii)
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.