Math, asked by shuvamdebnath6, 6 months ago

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

Answered by anjali5087
22

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

Answered by Anonymous
2

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

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