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Question-29048




Question Number 29048 by $@ty@m last updated on 03/Feb/18
Commented by ajfour last updated on 04/Feb/18
thanks for verifying Sir!
$${thanks}\:{for}\:{verifying}\:{Sir}! \\ $$
Commented by $@ty@m last updated on 04/Feb/18
Sorry  I dont have the answer.
$${Sorry} \\ $$$${I}\:{dont}\:{have}\:{the}\:{answer}. \\ $$
Commented by ajfour last updated on 04/Feb/18
please explain Sir!  i think something wrong with  question..
$${please}\:{explain}\:{Sir}! \\ $$$${i}\:{think}\:{something}\:{wrong}\:{with} \\ $$$${question}.. \\ $$
Commented by ajfour last updated on 04/Feb/18
S_1 =S_2  =A    (say)  S_1 +S_2 =S_3 +S_4  = 2A    ...(i)  3(S_1 +S_3 )=S_2 +S_4   ⇒   3A+3S_3 =A+S_4   ⇒    3S_3 +2A=S_4   using (i)            3S_3 +S_3 +S_4 =S_4   ⇒       S_3 =0   .  ⇒     OD =0 .  wrong data in question !
$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \:={A}\:\:\:\:\left({say}\right) \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={S}_{\mathrm{3}} +{S}_{\mathrm{4}} \:=\:\mathrm{2}{A}\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{3}\left({S}_{\mathrm{1}} +{S}_{\mathrm{3}} \right)={S}_{\mathrm{2}} +{S}_{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\mathrm{3}{A}+\mathrm{3}{S}_{\mathrm{3}} ={A}+{S}_{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{3}{S}_{\mathrm{3}} +\mathrm{2}{A}={S}_{\mathrm{4}} \\ $$$${using}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}{S}_{\mathrm{3}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} ={S}_{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:{S}_{\mathrm{3}} =\mathrm{0}\:\:\:. \\ $$$$\Rightarrow\:\:\:\:\:{OD}\:=\mathrm{0}\:. \\ $$$${wrong}\:{data}\:{in}\:{question}\:! \\ $$
Commented by ajfour last updated on 04/Feb/18
Commented by NECx last updated on 03/Feb/18
satyam...... ive not seen you  online for a long while. It feels  cool to see you  back here on the  forum.
$${satyam}……\:{ive}\:{not}\:{seen}\:{you} \\ $$$${online}\:{for}\:{a}\:{long}\:{while}.\:{It}\:{feels} \\ $$$${cool}\:{to}\:{see}\:{you}\:\:{back}\:{here}\:{on}\:{the} \\ $$$${forum}. \\ $$
Commented by $@ty@m last updated on 03/Feb/18
I open this app everyday but  most of the times I fail to solve   the prooblems.  However I am learning a lot on  this platform.
$${I}\:{open}\:{this}\:{app}\:{everyday}\:{but} \\ $$$${most}\:{of}\:{the}\:{times}\:{I}\:{fail}\:{to}\:{solve}\: \\ $$$${the}\:{prooblems}. \\ $$$${However}\:{I}\:{am}\:{learning}\:{a}\:{lot}\:{on} \\ $$$${this}\:{platform}. \\ $$
Commented by $@ty@m last updated on 03/Feb/18
in which class text book   can i find this theorem?
$${in}\:{which}\:{class}\:{text}\:{book}\: \\ $$$${can}\:{i}\:{find}\:{this}\:{theorem}? \\ $$
Commented by $@ty@m last updated on 03/Feb/18
Commented by $@ty@m last updated on 03/Feb/18
is it okay now?
$${is}\:{it}\:{okay}\:{now}? \\ $$
Commented by Tinkutara last updated on 03/Feb/18
BF=3. This is direct application of Ceva's Theorem.
Commented by Tinkutara last updated on 03/Feb/18
OD=4
Commented by Tinkutara last updated on 03/Feb/18
Commented by $@ty@m last updated on 03/Feb/18
Thanks a lot.
$${Thanks}\:{a}\:{lot}. \\ $$
Commented by Tinkutara last updated on 03/Feb/18
http://ibb.co/keD4G6
Commented by Tinkutara last updated on 04/Feb/18
OD=3 or 4? What is the answer? I am getting two different answers.
Commented by Tinkutara last updated on 04/Feb/18
This way it comes 3.
Commented by Tinkutara last updated on 04/Feb/18
Commented by mrW2 last updated on 04/Feb/18
you are right sir.  Following relations are valid:  DC=2BD  AF=2BF  AO=3OD
$${you}\:{are}\:{right}\:{sir}. \\ $$$${Following}\:{relations}\:{are}\:{valid}: \\ $$$${DC}=\mathrm{2}{BD} \\ $$$${AF}=\mathrm{2}{BF} \\ $$$${AO}=\mathrm{3}{OD} \\ $$
Answered by mrW2 last updated on 04/Feb/18
Commented by mrW2 last updated on 04/Feb/18
ΔCIE∼ΔCDA  ⇒CI=DI    ΔBOD∼ΔBEI  ⇒BD=DI  ⇒DC=2BD  therefore it′s not possible that  BD=5 and DC=15!    ΔAHE∼ΔAFC  ⇒AH=HF    ΔBFO∼ΔBHE  ⇒BF=FH  ⇒AF=2BF  or BF=((AF)/2)=(9/2)=4.5    OD=2IK  IK+KE=2OD  ⇒((OD)/2)+KE=2OD  ⇒KE=(3/2)OD  AO=2KE  ⇒AO=3OD  or OD=((AO)/3)=((12)/3)=4
$$\Delta{CIE}\sim\Delta{CDA} \\ $$$$\Rightarrow{CI}={DI} \\ $$$$ \\ $$$$\Delta{BOD}\sim\Delta{BEI} \\ $$$$\Rightarrow{BD}={DI} \\ $$$$\Rightarrow{DC}=\mathrm{2}{BD} \\ $$$${therefore}\:{it}'{s}\:{not}\:{possible}\:{that} \\ $$$${BD}=\mathrm{5}\:{and}\:{DC}=\mathrm{15}! \\ $$$$ \\ $$$$\Delta{AHE}\sim\Delta{AFC} \\ $$$$\Rightarrow{AH}={HF} \\ $$$$ \\ $$$$\Delta{BFO}\sim\Delta{BHE} \\ $$$$\Rightarrow{BF}={FH} \\ $$$$\Rightarrow{AF}=\mathrm{2}{BF} \\ $$$${or}\:{BF}=\frac{{AF}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{4}.\mathrm{5} \\ $$$$ \\ $$$${OD}=\mathrm{2}{IK} \\ $$$${IK}+{KE}=\mathrm{2}{OD} \\ $$$$\Rightarrow\frac{{OD}}{\mathrm{2}}+{KE}=\mathrm{2}{OD} \\ $$$$\Rightarrow{KE}=\frac{\mathrm{3}}{\mathrm{2}}{OD} \\ $$$${AO}=\mathrm{2}{KE} \\ $$$$\Rightarrow{AO}=\mathrm{3}{OD} \\ $$$${or}\:{OD}=\frac{{AO}}{\mathrm{3}}=\frac{\mathrm{12}}{\mathrm{3}}=\mathrm{4} \\ $$

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