Question

Triangles ABC and DBC are on the same base BC with A, D on opposite sides Of BC , such that ar ( ∆ ABC = ar ( ∆ DBC ) .Show that BC bisects AD.




Go ogle se copy maat karna .​

Answer 1

Explanation:

Answer

Given: Figure ABC and DBC are on the same base BC with vertices A and D or opposite sides of BC such that ar(△ABC)=ar(△DBC)

To Prove: BC bisects AD

Proof: ar(△ABC)=ar(△DBC)

⇒

2

1

×BC×AM

⇒

2

1

×BC×DN

Area of triangle =

2

1

× Base × Corresponding altitude

⇒ AM=DN−−−−−(1)

In △AMD and △DNO

AM=DN (prove (1))

∠AMN=∠DNO (each 90

o

)

∠AOM=∠DON

vertically opposite angles

∴ △AMO⊥△DNO

are to AAS congurence rule

∴ AD=DO(CPCT)

⇒ BC bisects AD

Answer 2

Answer:

Given: Figure ABC and DBC are on the same base BC with vertices A and D or opposite sides of BC such that ar(△ABC)=ar(△DBC)

To Prove: BC bisects AD

Proof: ar(△ABC)=ar(△DBC)

⇒

2

1

×BC×AM

⇒

2

1

×BC×DN

Area of triangle =

2

1

× Base × Corresponding altitude

⇒ AM=DN−−−−−(1)

In △AMD and △DNO

AM=DN (prove (1))

∠AMN=∠DNO (each 90

o

)

∠AOM=∠DON

vertically opposite angles

∴ △AMO⊥△DNO

are to AAS congurence rule

∴ AD=DO(CPCT)

⇒ BC bisects AD