ABC and DBC are two triangles on the same base BC.If AD intersects BC at O,show that ar(ABC)/ar(DBC)=AO/DO

const:draw AP and DQ perpendicular to BC
Proof:In Triangle AOP and DOQ,
Angle AOP=DOQ (VOP)
ANGLE APO=DQO=90(BY CONST.)
By AA similarity,AOP similar to DOQ
AP/DQ=AO/DO(CPST)----(1)
In Triangle ABC & DBC
ar(ABC)=1/2(BC*AP)----(2)
ar(DBC)=1/2(BC*DQ)----(3)
(2)/(3)
ar(ABC)/ar(DBC) =[1/2(BC*AP)]/[1/2(BC*DQ)]
ar(ABC)/ar(DBC) = AP/DQ----(4)
​ (1) in (4)
ar(ABC)/ar(DBC) = AO/DO

how to do it

in triangle ABC
ar[ABC]=1/2 base*height=1/2 BC*AO.....{1}
in triangle DBC
ar[DBC]=1/2 base*height=1/2 BC*OD......{2}

DIVIDING eq...{1} and {2}

ar[ABC]/ar[DBC]= 1/2 BC*AO/1/2 BC*OD
THEREFORE
ar[ABC]/ar[DBC]=AO/OD
HENCE PROVED!!
thumsup folks!

1 construct perpendiculars from vertices to the base
2 prove the triangles similar
3 then apply the rule of ratios of 2 similar triangles
4 apply thales theorem and equate it
5 and you are done!

figure?

const:draw AP and DQ perpendicular to BC Proof:In Triangle AOP and DOQ, Angle AOP=DOQ (VOP) ANGLE APO=DQO=90(BY CONST.) By AA similarity,AOP similar to DOQ =>AP/DQ=AO/DO ------>(1) ar(ABC)=1/2(BC*AP) and ar(DBC)=1/2(BC*DQ) ar(ABC)/ar(DBC)=AP/DQ FROM (1), ar(ABC)/ar(DBC)=AO/DO Hence Proved

1 construct perpendiculars from vertices to the base 2 prove the triangles similar 3 then apply the rule of ratios of 2 similar triangles 4 apply thales theorem and equate it 5 and you are done!

GIVEN-Triangle ABC and triangle DBC are on the same base
To Prove- Ar(ABC)/Ar(DBC) =AO/DO
Construction- Draw AP and DQ perpendicular to BC
Solution- In triangle AOP and triangle DOQ 
                  angle AOB  =  angle DOQ ( ver. opp.)
                  angle APO  =   angle  DQO = 90 degree
              => triangle AOP~ triangle DOQ (AA criteria)
              => AP/DQ = AO/DO   - (1)
                   
                    Ar (ABC)  =1/2 *BC*AP    - (2)
                    Ar(DBC)   = 1/2*BC*DQ    -(3)
On dividing 2 and 3, we get
                                             Ar(ABC)/Ar(DBC)   =  AP/DQ
​                                            => Ar(ABC) / Ar (DBC)  =  AO/DO      -  (from  1)
                                                                                                                                Hence proved
 

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