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.
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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
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