D is
any point on the side BC of triangle ABC. P and Q are centroids of triangle ABD and triangle ADO
respectively. Let us prove that PQ || BC.
Answers
Step-by-step explanation:
In△ABC,D is the mid-point of AB and D is any point on BC.
IfCQ∣∣PD meets AB in Q.
ThenIn△ABC,
we have to prove that BPQ=
2
1
ar(ABC)
Construct DC.
Since D is the mid point of AB in △BC,CD is the median.
ar(△BCD)=
2
1
ar(△ABC)−(i)
Since △PDQ&△PDC are in the same PD and between the same parallel lines PD&QC.
∴ar(△PDQ)=ar(△PDC)−(ii)
From(i)&(ii)
ar(△BCD)=
2
1
ar(△ABC)
ar(△BPD)+ar(△PDQ)=
2
1
ar(△ABC)
{areaof△PDC=PDQ}
⟹ar(△BPQ)=
2
1
ar(△ABC)
Hope it is helpful to you
Answer:
In triangle ABD,
AB+BD>AD [because, the sum of any two sides of a triangle is always greater than the third side]
In triangle ADC,
AC+DC>AD [because, the sum of any two sides of a triangle is always greater than the third side]
Adding these we get,
AB+BD+AC+DC>AD+AD
⇒AB+(BD+DC)+AC>2AD
⇒AB+BC+AC>2AD
Hence proved