Question

In triangle ABC,D is midpoint of AB.E is any point of BC.CF II ED meets AB in F.Show that area of triangle BEF = 1/2 area of triangle ABC.

Answer 1

In ABC, D is the mid-point ofAB and P is any point on BC. If CQ || PD me...

In triangle ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q then prove that ar (BPQ) =1/2ar (ABC)

Construction :- Join DC

To prove :- Since D is the mid-point of AB. so, in triangle ABC, CD is the median.

ar(triangle BCD = 1/2 ar (triangle ABC) ..... (1)

Since, triangle PDQ and triangle PDC are on the same base PD and between the same parallels lines PD QC.

therefore, ar(triangle PDQ) = ar(triangle PDC) ................ (2)

From (1) and (2)

ar(triangle BCD) = 1/2 ar(triangle ABC)

= ar(triangle BPD) + ar(triangle PDC) = 1/2 ar(triangle ABC)

= ar(triangle BPD) + ar(triangle PDQ) = 1/2 ar(triangle ABC) { area of triangle PDC = PDQ}

= ar(triangleBPQ) = 1/2 ar(triangle ABC)

Hence Proved ar(triangleBPQ) = 1/2 ar(triangle ABC)

i hope it helps

Answer 2

Answer:

Step-by-step explanation:

In ABC, D is the mid-point ofAB and P is any point on BC. If CQ || PD me...

In triangle ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q then prove that ar (BPQ) =1/2ar (ABC)

Construction :- Join DC

To prove :- Since D is the mid-point of AB. so, in triangle ABC, CD is the median.

ar(triangle BCD = 1/2 ar (triangle ABC) ..... (1)

Since, triangle PDQ and triangle PDC are on the same base PD and between the same parallels lines PD QC.

therefore, ar(triangle PDQ) = ar(triangle PDC) ................ (2)

From (1) and (2)

ar(triangle BCD) = 1/2 ar(triangle ABC)

= ar(triangle BPD) + ar(triangle PDC) = 1/2 ar(triangle ABC)

= ar(triangle BPD) + ar(triangle PDQ) = 1/2 ar(triangle ABC) { area of triangle PDC = PDQ}

= ar(triangleBPQ) = 1/2 ar(triangle ABC)

Hence Proved ar(triangleBPQ) = 1/2 ar(triangle ABC)

i hope it helps