# Proove that the diagnols of an isosceles trapezium are equal in heightdetailed proof

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Considering the isosceles trapezium PQRS, whose parallel sides are PS and QR. The base of isosceles trapezium PQRS is PS.

You need to prove that the length of diagonals PR and QS of isosceles trapezium PQRS are equal, hence, you need to prove that the triangles PQR and QRS are congruent.

In triangles PQR and QRS, the length of the sides PQ and RS are equal, because the non-parallel sides of isosceles trapezium are equal.

The side QR is common to both triangles and the angles hat(PQR) and hat(QRS) are equal.

Hence, since the pair of corresponding sides and the included angles are equal, yields that the triangles PQR and QRS are congruent (Side Angle Side).

Since the triangles PQR and QRS are congruent, hence, the sides PR and QS are also equal.

**Hence, using the congrunce of triangles PQR and QRS yields that the diagonals PR and QS of isosceles trapezium PQRS are equal.**

An isoceles trapezium is a trapezium whose in which the pair of non parallel side are equal. Let us consider an isosceles trpezium ABCD where AB & DC are parallel sides and BC & AD are non parallel sides, then BC = AD. The lines AC and BD are the two diagonals of the trapezium. Let us draw perpendiculars AE and BF from points A and B on the side CD respectively. AE and BF will be the height of the diagonal AC and BD respectively. We are require to prove that diagonals of this trapezium are of equal in height i.e. we are require to prove that AE = BF

Proof : Let us take two right angled triangles BCF and AED .

The side BC = side AD [ Given ABCD is an isosceles trapezium ]

Angle BCF = Angle ADE [ Since side BC = side AE ]

Angle BFC = Angle AED = 90 degrees [ BF and AE are perpendiculars on CD]

Triangle BCF is congruent to triangle AED [ test A.A.S ]

There the side BF = side AE [ corresponding sides of congruent triangles will also be equal ]

**Proved**

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