In the adjoining figure, `AP ⊥ BC` ,AD||BC ,then find `A(Delta ABC):A(Delta BCD)` .

In adjoining figure, AP _|_ BC , AD || BC ,then find Ar.(Delta ABC) : Ar.(Delta BCD) Fig.

In the adjoining figure, AP bot BC, AD abs() BC , then find A(DeltaABC):A(DeltaBCD) .

In the following figure AP_|_BC, AD||BC, then find A(DeltaABC):A(DeltaBCD) .

In the adjoining figure, BC bot AB, AD bot AB, BC = 4, AD = 8, then find (A(DeltaABC))/(A(DeltaADB)) ..

In the adjoining figure, DE || BC . Prove that area (Delta ACD) = " area " (Delta ABE)

In the adjoining figure,E is a point on the median AD of a Delta ABC. Show that ar (Delta ABE)=(Delta ACE)

In the adjoining figure, D is the mid-point of side AB of Delta ABC and P be any point on side BC. If CQ ||PD , then prove that: area of Delta BPQ = (1)/(2) xx " area of " Delta ABC

In the adjoining figure, AD is the median of Delta ABC and X be any point on side AD. Prove that: area (Delta ABX) = " area " (Delta ACX)

In the given figure, if DE |\| BC, then find the ratio of area of (triangle ADE) " and " (squareDECB).

In the adjoining figure D, E and F are the mid-points of the sides BC, AC and AB respectively. triangle DEF is congruent to triangle :