In `DeltaABC`, AD divides BC in thr ratio `m:n`. Show that `( "Area" (DeltaABD))/( "Area" (DeltaADC))=(m)/(n)`

In the given figure, AD is median of DeltaABC and E is any point on median AD. Prove that Area (DeltaABE) = Area (DeltaACE)

In the following figure, DE is parallel to BC. Show that : (i) "Area" (DeltaADC)= "Area" (DeltaAEB) (ii) "Area" (DeltaBOD)= "Area" (DeltaCOE) .

In the figure of question 2 , if E is the mid point of median AD, then prove that : "Area" (DeltaABE)=(1)/(4) "Area" (DeltaABC) .

In the figure of question 2 , if E is the mid point of median AD, then prove that : "Area" (DeltaABE)=(1)/(4) "Area" (DeltaABC) .

In a DeltaABC, D in mid-point of BC, AD is produced upto E so that DE=AD , prove that : (i) DeltaABD and DeltaECD are congruent (ii) AB=EC (iii) AB is parallel to EC.

In the given figure, DE is parallel to the base BC of triangle ABC and AD: DB = 5: 4. Find the ratio : ("Area of "DeltaDEF)/("Area of"Delta BFC)

The base BC of triangle ABC is divided at D so that BD = (1)/(2) DC. Prove that area of DeltaABD=(1)/(3) of the area of DeltaABC .

In the given figure, diagonals PR and QS of thr parallelogram PQRS intersect at point O and LM is parallel to PS. Show that : (i) 2 "Area" (DeltaPOS)= "Area" (////gmPMLS) (ii) "Area" (DeltaPOS)+ "Area" (DeltaQOR) =(1)/(2) "Area" (////gmPQRS) (iii) "Area" (DeltaPOS)+ "Area" (DeltaQOR) = "Area" (DeltaPOQ)+ "Area" (tDeltaSOR)

ABCD is a trapezium with AB////DC . A line parallel to AC intersects Ab at point M and BC at point N. Prove that : "area of" DeltaADM= "area of" DeltaACN .

In DeltaABC , a point P on BC divided BC in the ratio 1:1. what is the line segment joining vertex A and P called ?