Q11 of 83 Page 351

In the adjoining figure, D and E are points on the sides AB and AC of ∆ABC such that ar(∆BCE) = ar(∆BCD).

Show that DE ‖ BC.


Given


A triangle ABC in which points D and E lie on AB and AC of ∆ABC such that ar(∆BCE) = ar(∆BCD).


To prove: DE ‖ BC


Proof:


Here, from the figure we know that BCE and BCD lie on same base BC and


It is given that area(∆BCE) = area(∆BCD)


Since two triangle have same base and same area they should equal altitude(height)


That means they lie between two parallel lines


That is DE ‖ BC


DE ‖ BC


Hence proved


More from this chapter

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9

In the adjoining figure, ABCD is a trapezium in which AB ‖ DC and its diagonals AC and BD intersect at O. Prove that ar(∆AOD) = ar(∆BOC).

 10

In the adjoining figure, DE ‖ BC. Prove that

(i) ar(∆ACD) = ar(∆ABE),


(ii) ar(∆OCE) = ar(∆OBD).


 12

In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that

(i) ar(∆OAB) + ar(∆OCD) = ar(‖gm ABCD),


(ii)ar(∆OAD) + ar(∆OBC) = ar(‖gm ABCD).


 13

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.

Prove that ar(∆ABP) = (quad.ABCD).