In a trapezium ABCD, AB is parallel to DC and L is the midpoint of BC. Through L, a line LM is drawn parallel to AD which meets AB in X and DC produced in Y. Prove that area of ABCD= area of AXYD

hmm...

Hmmmmmmmm xdxdxdxdxdxd

aspki kitni clasess hoti hai

bolo

6 classes and 1 break

Hii,
 

 

Given: ABCD is a trapezium with AB || CD

L is the mid point of BC and PQ || AD

 

Now, In ∆CLQ and ∆BLP

∠LCQ = ∠LBP        (∵ AB || CDQ and BC is the transversal so, alternate interior opposite angles)

CL = BL                 (∵ L is the mid point of BC)

∠CLQ = ∠BLP           (Vertically opposite angles)

So ∆CLQ ≅ ∆BLP        ( by ASA congruency criteria)

⇒ Area (∆CLQ) = Area (∆BLP)

⇒ Area (ABCLP) + Area (∆CLQ) = Area (ADCLP) + Area (∆BLP)

 

Hence Area (APQD) = Area (ABCD)

Regards

okkk....

bhavika ab aap hi batao ki mai kya kro....priya ne mera dimag kharab krr diya hai

sorryy....bolne ke liye....prr aap shantanu ko pareshan kyu krr rhe hoo

Ok Varun sorry but such kind of language is not allowed on Ask n Ans Forum na... Why don't you understand? I didn't say anything to anyone just I told bad words aren't allowed

Varun because he is ignoring me when I reply on his post that's why. I am sorry to you but not to him.