Question

In ADC, E and B are the points on the sides AD and AC respectively such that ∠ABE=∠ADC. If AE = 6 cm,BC = 2 cm, BE = 3 cm and CD = 5 cm, then (AB + DE)is equal to:

Answer 1

Answer:

If a line intersect the two sides in distinct points and the two sides are divided in the same ratio then the line drawn is parallel to the third side of the triangle.

EC

AE

=

9

6

=

3

2

.......(1)

DB

AD

=

12

8

=

3

2

......(2)

As (1)=(2)

So ED∥CB

As ED∥CB

∠AED=∠ACB

∠ADE=∠ABC

So by AAA △AED∼△ACB

Hence

AC

AE

=

CB

ED

AE+EC

AE

=

CB

ED

[∴AC=AE+EC]

6+9

6

=

CB

ED

15

6

=

5

2

=

CB

ED

∴CB=

2

5

ED [henceproved]