Question

In quadrilateral ABCD, if the mid points of one pair of opposite sides and the point of intersection of the diagonals are collinear, using vector methods, prove that the quadrilateral ABCD is a trapezium.

Answer 1

Step-by-step explanation:

Given:

∠BDC=90

∘

∠ABC=90

∘

Let ∠BCD=x

∘

Using triangle sum property in △BDC, ∠DBC=90−x

∘

Also ∠ABD=x

∘

Using triangle sum property in △ADB, ∠BAD=90−x

∘

Now considering △BDC and △ADB

∠BDC=∠BDA [∵∠BDC=∠BDA=90

∘

]

∠DBC=∠DAB [∵∠DBC=∠DAB=(90−x)

∘

]

So by AA

△BDC∼△ADB

Hence

CD

BD

=

BD

AD

CD

8

=

8

4

CD=16

Hence CD=16cm