*P* is any point on base *BC* of Δ*ABC** *and *D* is the mid-point of* BC*. *DE* is drawn parallel to*PA* to meet *AC* at *E*. If ar (Δ*ABC*) = 12 cm^{2}, then find area of Δ*EPC*.

#### Solution

**Given:** Area (ABC) = 12 cm^{2}, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.

Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,

Area (APE) = Area (APD)

⇒ Area (APM) + Area (AME) = Area (APM) + Area (PMD)

⇒ Area (AME) = Area (PMD) …… (1)

Since, median divide triangles into two equal parts. So,

Area (ADC) = `1/2` Area (ABC) = `12/2` = 6 cm^{2}

⇒ Area (ADC) = Area (MDCE) + Area (AME)

⇒Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1))

⇒ Area (ADC) = Area (PEC)

Therefore,

Area (PEC) = 6 cm^{2}.