Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
Solution:
Given, ABCD is a trapezium
M and N are the midpoints of parallel sides AB and DC of the trapezium.
The line segment joining M and N is perpendicular to both the sides AB and DC.
We have to prove that AD = BC
Since M is the midpoint of AB
AM = MB
Considering triangles AMN and BMN,
AM = MB
∠3 = ∠4 = 90
Common side = MN
By SAS criterion, the triangles AMN and BMN are congruent
By CPCTC, ∠1 = ∠2
Multiplying by -1 on both sides,
(-1)∠1 = (-1)∠2
Adding 90 on both sides,
90° - ∠1 = 90° - ∠2
From the figure,
∠AND = ∠BNC
Considering triangle ADN and BCN,
Since triangles AMN and BMN are congruent, AN = BN
∠AND = ∠BNC
Since N is the midpoint of CD
CN = DN
By SAS criterion, the triangles AMN and BMN are congruent.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are congruent, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
AD = BC
Therefore, it is proved that AD = BC
✦ Try This: In right angled triangle ABC, right angled at C,M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM=CM. Point D is joined to point B. Show that: CM = 1/2 AB
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 16
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC
Summary:
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. It is proven that AD = BC by SAS criterion
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