Question

In a Δ ABC, if L and M are points on AB and AC respectively such that LM||BC. Prove that:
(i) ar(Δ LCM) = ar(Δ LBM)
(ii) ar(Δ LBC) = ar(Δ MBC)
(iii) ar(Δ ABM) = ar(Δ ACL)
(iv) ar(Δ LOB) = ar(Δ MOC)

Answer 1

Given : In a Δ ABC, if L and M are points on AB and AC respectively such that LM || BC.

 

 

To prove : (i) ar(Δ LCM) = ar(Δ LBM)

(ii) ar(Δ LBC) = ar(Δ MBC)

(iii) ar(Δ ABM) = ar(Δ ACL)

(iv) ar(Δ LOB) = ar(Δ MOC)

 

 

Proof :  

(i) Clearly, ∆'s LMB and LMC are on the same base LM and between the same parallels LM and BC.

∴ ar(Δ LCM) = ar(Δ LBM) ……….. (i)

 

(ii) Here, ∆'s LBC and MBC are on the same base BC and between the same parallels LM and BC.

∴ ar (ΔLBC) = ar (ΔMBC) ……...(ii)

 

(iii) We have,ar (∆LMB)  = ar (∆LMC) [From eq (i)]

On adding ar (∆ALM) on both sides :  

ar (∆LMB)  + ar (∆ALM) = ar (∆LMC) + ar (∆ALM)

ar (∆ABM)  = ar (∆ACL)

 

(iv) We have,

ar (ΔLBC) = ar (ΔMBC)

[From eq (ii)]

On subtracting ar (∆BOC) from both sides :  

ar (ΔLBC) - ar (∆BOC) = ar (ΔMBC) - ar (∆BOC)

ar (∆LOB) = ar(∆MOC)

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answer 2

Answer:

since, ΔBCA and ΔBYA are on the same base BA and between asme parallels BA and CY Then area (ΔBCA)

\bf \red{→}ar(BYA)→ar(BYA)

\bf \pink{→}ar(Δ CBX) + ar(Δ BXA) =→ar(ΔCBX)+ar(ΔBXA)=

\bf = ar(Δ BXA) + ar(Δ AXY)=ar(ΔBXA)+ar(ΔAXY)

\bf \red{→}ar(Δ CBX) = ar(ΔBXA) =→ar(ΔCBX)=ar(ΔBXA)=

\bf = ar(Δ BXA) + ar(Δ AXY)=ar(ΔBXA)+ar(ΔAXY)

\bf \pink→ ar (Δ CBX )= ar(Δ AXY) \: \: ...(i)→ar(ΔCBX)=ar(ΔAXY)...(i)

Since, ΔACE are on the same base AE and between same parallels CD and AE

Then, ar (ΔACE) ar (ΔADE)

= (ΔCLA) + ar (ΔAZE) = ar (ΔAZE) + ar (ΔDZE)

= ar (ΔCZA) = (ΔDZE) ...... (2)

Since, ΔCBY and ΔCAY ar on the same base CY and between same parallels

BA and CY

Then ar (ΔCBY) ar (ΔCAY)

Adding ar (CYG) on both sides, we get

= ar (ΔCBX) + ar (ΔCYZ) = (ΔCAY) + ar (ΔCYZ)

= ar (BCZX) = ar (ΔCZA) .......... (3)

Compare equation (2) and (3) are (BCZY) = ar (ΔDZE)