Supposed ABCD is a rectangle, F and G are points on such that DF=GC, and AFE and BGE are lines. Prove that DE=EC.

Area of

Δ

A

E

B

=

12.5

units.

Explanation:

The above is described in the figure below, the difference being that we have drawn

F

K

⊥

A

B

, intersecting

C

D

at

H

.

enter image source here

Let

H

F

=

x

and

E

H

=

y

.

Now as

C

D

||

A

B

, we have similar triangles

E

F

H

and

E

A

K

and hence

H

F

A

K

=

E

H

E

K

i.e.

x

1

+

x

=

y

3

+

y

or

3

x

+

x

y

=

y

+

x

y

or

3

x

=

y

i.e.

x

=

y

3

Similarly in similar triangles

E

G

H

and

E

B

K

we have

H

G

B

K

=

E

H

E

K

i.e.

2

−

x

2

−

x

+

2

=

y

3

+

y

or

2

−

x

4

−

x

=

y

3

+

y

or

6

−

3

x

+

2

y

−

x

y

=

4

y

−

x

y

or

2

y

+

3

x

=

6

and as

x

=

y

3

2

y

+

y

=

6

or

3

y

=

6

i.e.

y

=

2

and

x

=

2

3

As such

E

K

=

E

H

+

H

K

=

2

+

3

=

5

and area of triangle

A

E

B

=

E

K

×

A

B

2

=

5

×

5

2

=

12.5

units

answer: 2 and 1

step-by-step explanation: got this question