ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig). Show that F is the mid-point of BC.
Let EF intersect DB at G.
By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
In ∆ABD,
EF || AB and E is the mid-point of AD.
Therefore, G will be the mid-point of DB.
As EF || AB and AB || CD,
∠EF || CD (Two lines parallel to the same line are parallel to each other)
In ∆BCD, GF || CD and G is the mid-point of line BD.
Therefore, by using converse of mid-point theorem, F is the mid-point of BC.
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.