Question

In triangle ABC ,D is the midpoint of AB. P IS ANY POINT OF BC. CQ || PD meets AB in Q . Show that ar ( ∆BPQ) = ½(∆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) = 1/2ar(triangle ABC)
= ar(triangle BPD) + ar(triangle PDQ) = 1/2ar(triangle ABC) { area of triangle PDC = PDQ}
= ar(triangleBPQ) = 1/2 ar(triangle ABC)