Question

1. ABC and ABD are two triangles on the same base AB. If the line segment CD is bisected by Ab at O, show that ar(ΔABC) = ar(ΔABD)

2. In a triangle ABC, E is the mid-point of median AD. Show that ar(ΔBED) = 1/4ar(ΔABC).

Answer 1

Given: ∆ABC and ∆ ABD on the same base AB

and AB bisects CD, i.e. , OC = OD

To proove: ar (ABC) = ar ( ABD)

Proof:

In ∆ACD,

Since OC = OD

: OA in the Median.

= ar ( ∆AOC) = ar (∆AOD)

Similarly , In ∆BCD

Since OC = OD

:: OB is the median

= ar (∆BOC) = ar ( ∆BOD) you

2.

Given: ∆ABC

with AD as medien i.e BD = CD &

E is the mid-point of AD, i.e. , AE = DE

To proove: ar ( BED) = 1/4 ar (ABC).

Proof:

AD is the median of ∆ ABC &

medien divides a triangle into two triangles of equal areas

:: ar (ABD) = ar (ACD)

== ar (ABD) = 1/2 ar (ABC)

Hope it's helpful for you ✌️✌️

Answer 2

1.

In triangle ABC, AO is the median (CD is bisected by AB at O)

So, ar(AOC)=ar(AOD)..........(i)

Also,

triangle BCD,BO is the median. (CD is bisected by AB at O)

So, ar(BOC) = ar(BOD)..........(ii)

Adding (i) and (ii),

We get,

ar(AOC)+ar(BOC)=ar(AOD)+(BOD)

⇒ ar(ABC) = ar(ABD)

Hence showed.

2.

AD is the median of Triangle ABC.Therefore, it will divide triangle ABC into two triangles of equal areas.

Area (triangle ABD)=Area ( triangle ACD)

⇒ Area (triangle ABD)=1/2 Area ( triangle ABC).....(i)

In triangle ABD, E is the mid point of AD.

Therefore, BE is the median.

Area (triangle BED)=Area ( triangle ABF)

⇒ Area (triangle BED)=1/2 Area ( triangle ABD)

⇒ Area (triangle BED)=1/2*1/3 Area ( triangle ABC).....from equation (i)

⇒ Area (triangle BED)=1/4 Area ( triangle ABC)

Hence showed.