Question-210235

Question Number 210235 by peter frank last updated on 03/Aug/24

Commented by Pnk2024 last updated on 03/Aug/24

△ADB ∼△EFB .... (A−A test) ⇒ (y/x) = ((AB)/(EB)) ....... (C.S.S.T) again △BCA∼△EFA ⇒ ((AE)/(AB))= (x/z) ........ (C.S.S.T.) now ((AE)/(AB))×((AB)/(EB))= (x/z)×(y/x) ⇒ ((AE)/(EB))=(y/z) ⇒ ((AE)/(EB+AE))=(y/(y+z)) ⇒ ((AE)/(AB))=(y/(y+z)) .........(ii) but from above ((AE)/(AB))=(x/z) ∴ (y/(y+z))=(x/z) ⇒ yz=xy+xz dividing by xyz ⇒ ((yz)/(xyz))=((xy)/(xyz))+((xz)/(xyz)) ⇒ (1/x)=(1/z)+(1/y) this is proved

$$\bigtriangleup{ADB}\:\sim\bigtriangleup{EFB}\:\:\:\:….\:\left({A}−{A}\:{test}\right) \\ $$$$\Rightarrow\:\frac{{y}}{{x}}\:=\:\frac{{AB}}{{EB}}\:\:\:…….\:\left({C}.{S}.{S}.{T}\right) \\ $$$${again} \\ $$$$\bigtriangleup{BCA}\sim\bigtriangleup{EFA} \\ $$$$\Rightarrow\:\frac{{AE}}{{AB}}=\:\frac{{x}}{{z}}\:……..\:\left({C}.{S}.{S}.{T}.\right) \\ $$$$\:{now}\:\:\frac{{AE}}{{AB}}×\frac{{AB}}{{EB}}=\:\frac{{x}}{{z}}×\frac{{y}}{{x}} \\ $$$$\Rightarrow\:\frac{{AE}}{{EB}}=\frac{{y}}{{z}} \\ $$$$\Rightarrow\:\frac{{AE}}{{EB}+{AE}}=\frac{{y}}{{y}+{z}} \\ $$$$\Rightarrow\:\frac{{AE}}{{AB}}=\frac{{y}}{{y}+{z}}\:\:………\left({ii}\right) \\ $$$${but}\:{from}\:{above}\:\:\:\frac{{AE}}{{AB}}=\frac{{x}}{{z}} \\ $$$$\therefore\:\:\frac{{y}}{{y}+{z}}=\frac{{x}}{{z}}\:\: \\ $$$$\:\Rightarrow\:{yz}={xy}+{xz} \\ $$$$\:{dividing}\:\:{by}\:{xyz} \\ $$$$\Rightarrow\:\frac{{yz}}{{xyz}}=\frac{{xy}}{{xyz}}+\frac{{xz}}{{xyz}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{y}}\:\:\:{this}\:{is}\:{proved} \\ $$$$ \\ $$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by mr W last updated on 04/Aug/24

Commented by mr W last updated on 04/Aug/24

(x/z)=((a+b)/b)=1+(a/b) ⇒(a/b)=(x/z)−1 (y/z)=((a+b)/a)=1+(b/a) ⇒(b/a)=(y/z)−1 ⇒((x/z)−1)((y/z)−1)=1 ⇒xy=(x+y)z ⇒(1/z)=(1/x)+(1/y) ✓

$$\frac{{x}}{{z}}=\frac{{a}+{b}}{{b}}=\mathrm{1}+\frac{{a}}{{b}}\:\Rightarrow\frac{{a}}{{b}}=\frac{{x}}{{z}}−\mathrm{1} \\ $$$$\frac{{y}}{{z}}=\frac{{a}+{b}}{{a}}=\mathrm{1}+\frac{{b}}{{a}}\:\Rightarrow\frac{{b}}{{a}}=\frac{{y}}{{z}}−\mathrm{1} \\ $$$$\Rightarrow\left(\frac{{x}}{{z}}−\mathrm{1}\right)\left(\frac{{y}}{{z}}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow{xy}=\left({x}+{y}\right){z} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:\checkmark \\ $$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by A5T last updated on 04/Aug/24

Commented by A5T last updated on 04/Aug/24

(x/y)=((BE)/(BA));(x/z)=((EA)/(BA)) ⇒(x/y)+(x/z)=((BE+EA)/(BA))=1⇒(1/y)+(1/z)=(1/x)

$$\frac{{x}}{{y}}=\frac{{BE}}{{BA}};\frac{{x}}{{z}}=\frac{{EA}}{{BA}} \\ $$$$\Rightarrow\frac{{x}}{{y}}+\frac{{x}}{{z}}=\frac{{BE}+{EA}}{{BA}}=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{x}} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply