Geometry: Similar triangles :- Prove that 1/OA + 1/OB = 2/ OC

[@Angel-gt2xh]

Best explanation. Thnx sir😄

[@JatheenVarmaBuddharaju]

Thank you beta. Please share. Wish you all the best.

[@kamaliniv]

thank you sir

[@ARMY-vi3zy]

Sir in last why we have devided by OA.OB.OC

[@JatheenVarmaBuddharaju]

@9:33, When we divide by OA * OB * OC each term on both sides we get what we need to prove. Try to divided each term by OA OB OC and see. Due to space I skipped a step there. Hope it helps :)

[@ARMY-vi3zy]

Ok sir

[@mahendrachoudhary3123]

Tq sir

[@Harshexplain95]

Mast explain karte ho aap sir.👃👃👃👃👃

[@gurleengura1893]

Sir is there any another method to do this question

[@JatheenVarmaBuddharaju]

Harkirat Gura, right now I am not aware of, but may be possible. Thanks for asking. Wish you all the best!!! Please subscribe and share: )

[@Prajapatikaran628]

Pagal

[@Prajapatikaran628]

Fact its

[@Prajapatikaran628]

Any one method

[@rpsingh4]

A dash divide BE.

[@JatheenVarmaBuddharaju]

?????? I did not get your query. Can you please be more clear if you have any doubts.

[@rpsingh4]

@@JatheenVarmaBuddharaju There must be a point b/w BE namely A' that divide BE.

[@JatheenVarmaBuddharaju]

Thanks for the reply. Why do you think so? Can you explain in detail. I want to know your thought process. We have 2 pairs of similar triangles and that should be enough to solve the problem.

[@rpsingh4]

@@JatheenVarmaBuddharaju if you add A' then line OA' is going to bisect line C' and C' E. Where C' is the point on the intersection of line AB and FE. Also lines BD and PR(love) make a parallelogram.

[@JatheenVarmaBuddharaju]

@@rpsingh4, ohh.. You are giving an alternative solution. Thank you, I will go through that. I thought you have a doubt in the given solution. Thanks for quick reply. Wish you all the best :)