Question

the point D divides the side BC of triangle ABc in the ratio m:n prove that ar(triangle ABD):ar(triangle ADC)=m:n

Answer 1

ar(Δ ABD) = ½*B * H

= ½ * BD * AE

ar(Δ ADC) = ½ * B * H

= ½ * CD * AE

Now, their ratio :

½*BD*AE/1/2*CD*AE = BD/CD

Therefore BD/CD = BD : CD i.e m:n

Hence proved.

 

I hope it helps……….

Answer 2

Given - ABC is a triangle and D divided BC into M:N Construction- draw AE perpendicular to BC Proof - let BD =MX and DC = NX At(ABD)= 1/2 B XH = 1/2 X BDXAE=1/2 X AE X MX.........1 AR(ADC)=1/2 B H =1/2XDCXAE=1/2AE X NX...........2 FROM 1 AND 2 WE GET ar(triangle ABD):ar(triangle ADC)=m:n