Q7 of 31 Page 232

E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F. Joined D, F. Let’s prove that

(i) triangle ADF = triangle ABE


(ii) triangle DEF = triangle BEC.

Given.


E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F


Formula used.


If 2 triangles are on same base And lies between 2 parallel lines then both triangles are equal.


If triangle and parallelogram are on same base And lies between 2 parallel lines then triangle gets half of parallelogram.


Draw the diagonal BD of parallelogram ABCD



In triangle ADF and triangle ADB


Both lies on same base AD


And AD || BC ∵ opposite sides of parallelogram are parallel


As BF is extended part of BC


Then AD || BF


∴ triangle ADF = triangle ADB = of parallelogram ABCD


∵ Diagonal of parallelogram gives 2 congruent triangles


In triangle ABE and parallelogram ABCD


Both lies on same base AB


And AB || DC ∵ opposite sides of parallelogram are parallel


∴ triangle ABE = of parallelogram ABCD


Both triangle ADF and triangle ABE are half of parallelogram


∴ triangle ADF = triangle ABE


If ∴ triangle ABE = of parallelogram ABCD


⇒ triangle ADE + triangle EBC = Another of parallelogram ABCD


∴ triangle ADF = of parallelogram ABCD


⇒ triangle ADF= triangle ADE+ triangle DEF


→ triangle ADE+ triangle DEF = triangle ADE+ triangle ABE


∴ triangle DEF = triangle ABE


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