Answer:
The area of ABCD is 36 cm².
Step-by-step explanation:
Given information: ABCD is a parallelogram, E is the midpoint of BC and AE ∩ BD=F.
In triangle EBF and ADF
[tex]\angle FBE=\angle FDA[/tex] (Alternate interior angles)
[tex]\angle FEB=\angle FAD[/tex] (Alternate interior angles)
[tex]\angle BFE=\angle DFA[/tex] (Vertically opposite angles)
Therefore triangle EBF and ADF are similar triangles by AA rule.
The sides BC and AD are opposite sides of parallelogram, therefore their lengths are equal. E is midpoint of BC.
[tex]\frac{BC}{AD}=\frac{1}{2}[/tex]
If a point divides the side of a triangle in m:n, then the line segment between the point and the opposite vertex, divides the area of triangle is m:n.
Therefore the sides of triangle are in proportion of 1:2 and we can say that the point F divides the line EA and BD in 1:2.
In triangle ABE, the line BF divides the area of triangle is 1:2.
[tex]\text{ Area of }\triangle BEF:\text{ Area of }\triangle ABF=1:2[/tex]
[tex]\text{ Area of }\triangle ABF=2\text{ Area of }\triangle BEF[/tex]
[tex]A_{\triangle ABF}=2A_{\triangle BEF}[/tex]
[tex]A_{\triangle ABF}=2\times 3=6[/tex]
In triangle ABD, the line FA divides the area of triangle is 1:2.
[tex]\text{ Area of }\triangle ABF:\text{ Area of }\triangle DFA=1:2[/tex]
[tex]\text{ Area of }\triangle DFA=2\text{ Area of }\triangle ABF[/tex]
[tex]A_{\triangle DFA}=2A_{\triangle ABF}[/tex]
[tex]A_{\triangle DFA}=2\times 6=12[/tex]
[tex]A_{\triangle ABD}=A_{\triangle ABF}+A_{\triangle ADF}[/tex]
[tex]A_{\triangle ABD}=6+12=18[/tex]
Since AD is a diagonal of the parallelogram, therefore AD divides the area of parallelogram in two equal parts.
[tex]A_{\triangle ABCD}=2A_{\triangle ABD}[/tex]
[tex]A_{\triangle ABCD}=2\times 18=36[/tex]
Therefore the area of ABCD is 36 cm².
