Prove that diagnols of an isosceles trepezium are equal in height

With the diagram

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You need to consider the trapezoid ABCD and you need to project the vertices C and D to the base of trapezoid AB.

The projection of vertex C falls in the point M and the projection of vertex D falls in the point N.

You should consider the right triangles AND and BMC

The problem provides the information that the trapezoid ABCD is isosceles, hence the lengths of non parallel sides AD and BC are equal and the angles made by AD and BC to the base AB are equal.

You need to prove that the heights of diagonals are equal, hence you need to prove that the lengths of sides CM and DN are equal.

The sides CM and DN are opposite to the equal angles, hence you may use sine function to prove that CM and DN are equal.

`sin AND = (ND)/(AD)`

`sin BMC = (CM)/(BC)`

Since the angles AND and BMC are equal => `sin AND = sin BMC` .

If `sin AND = sin BMC =gt (ND)/(AD) = (CM)/(BC)`

Since `AD=BC` and `(ND)/(AD) = (CM)/(BC) =gt CM = ND` .

**Hence, the last line proves that the height of diagonals in isosceles trapezium are equal.**

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