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Question Number 176531 by a.lgnaoui last updated on 20/Sep/22

Dans la figure ci−joint  AB∣∣ A^′ B^′    AA^′ =BB^′ =CC^′   Determiner le rapport       ((aire Δ(A^′ B^′ C^′ ))/(aire Δ(ABC)))=?

$${Dans}\:{la}\:{figure}\:{ci}−{joint} \\ $$$${AB}\mid\mid\:{A}^{'} {B}^{'} \:\:\:{AA}^{'} ={BB}^{'} ={CC}^{'} \\ $$$${Determiner}\:{le}\:{rapport} \\ $$$$\:\:\:\:\:\frac{{aire}\:\Delta\left({A}^{'} {B}^{'} {C}^{'} \right)}{{aire}\:\Delta\left({ABC}\right)}=? \\ $$

Commented by a.lgnaoui last updated on 20/Sep/22

Commented by a.lgnaoui last updated on 20/Sep/22

AA^′   BB^′   CC^′ Bissectrice de: ∡(A  ; B ; C)

$${AA}^{'} \:\:{BB}^{'} \:\:{CC}^{'} {Bissectrice}\:{de}:\:\measuredangle\left({A}\:\:;\:{B}\:;\:{C}\right) \\ $$

Commented by mr W last updated on 20/Sep/22

with given conditions we can only  find out that ∠A=∠B, i.e. AC=BC,  but we can not determine ((ΔA′B′C′)/(ΔABC))  uniquely.

$${with}\:{given}\:{conditions}\:{we}\:{can}\:{only} \\ $$$${find}\:{out}\:{that}\:\angle{A}=\angle{B},\:{i}.{e}.\:{AC}={BC}, \\ $$$${but}\:{we}\:{can}\:{not}\:{determine}\:\frac{\Delta{A}'{B}'{C}'}{\Delta{ABC}} \\ $$$${uniquely}. \\ $$

Commented by a.lgnaoui last updated on 20/Sep/22

exactly

$${exactly} \\ $$

Answered by RahulRajpoot last updated on 20/Sep/22

$$ \\ $$

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