Question

In an adjoining figure, the point D divides the side BC of Triangle ABC in the ratio m: n. Prove that Area of triangle ABC: Area triangle ABD = m: n. *In a figure X and Y are the midpoints of AC and AB respectively, QP parallel BC and CYQ and BXP are straight lines. Prove that area triangle ABP = area triangle ACQ. *In a figure, BD parallel CA, E is the midpoint of CA and BD = � CA. Prove that are triangle ABC =2 x area of triangle DBC. plz need answers urgently

Answer 1

according t ur first question related to m:n,
pls note we cn use BPT(Basic proportionality theorum) fr this..
mark the point on BC that divides in the ratio m:n as 'D'
draw a parellel line from D  that is parellel to AC.and mark the intersection at AB as 'E'
in triangle BED and BAC we notice that
now angle BDE=Angle BCA()angles formed by parellel lines DE and AC and transversal BC
and angle BED= angle BAC
common angle B
therefore both triagles r similar, therefore by BPT we prove that area of triangle BED:Area of trangle BAC=(m*m):(n*n)

i realised ur last question's 1 part contains unicode character, pls correct this:)