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If-in-ABC-r-1-r-2-r-3-r-prove-that-triangle-is-right-angled-

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Question Number 16273 by Tinkutara last updated on 20/Jun/17
If in ΔABC r_1 = r_2 + r_3 + r, prove that triangle is right angled.
$$\mathrm{If}\:\mathrm{in}\:\Delta{ABC}\:{r}_{\mathrm{1}} \:=\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:+\:{r},\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled}. \\ $$
Commented by mrW1 last updated on 20/Jun/17
what is r,r_1 ,...?
$$\mathrm{what}\:\mathrm{is}\:\mathrm{r},\mathrm{r}_{\mathrm{1}} ,…? \\ $$
Commented by Tinkutara last updated on 20/Jun/17
r is the inradius of ΔABC. r_1 , r_2 and r_3 are the exradii of ΔABC opposite to angles A, B and C respectively.
$${r}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inradius}\:\mathrm{of}\:\Delta{ABC}. \\ $$$${r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} \:\mathrm{and}\:{r}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{exradii}\:\mathrm{of}\:\Delta{ABC} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{angles}\:{A},\:{B}\:\mathrm{and}\:{C} \\ $$$$\mathrm{respectively}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17
p.tg(A/2)=p.tg(B/2)+p.tg(C/2)+(S/p) ⇒tg(A/2)=tg(B/2)+tg(C/2)+(S/p^2 ) tg(A/2).tg(B/2).tg(C/2)=(S/p^2 ) ⇒tg(A/2)=tg(B/2)+tg(C/2)+tg(A/2).tg(B/2).tg(C/2) ⇒tg(A/2)(1−tg(B/2).tg(C/2))=tg(B/2)+tg(C/2) ⇒tg(A/2)=((tg(B/2)+tg(C/2))/(1−tg(B/2).tg(C/2)))=tg((B/2)+(C/2))⇒ (A/2)=((B+C)/2)⇒A=B+C=180−A⇒A=90^• .■
$${p}.{tg}\frac{{A}}{\mathrm{2}}={p}.{tg}\frac{{B}}{\mathrm{2}}+{p}.{tg}\frac{{C}}{\mathrm{2}}+\frac{{S}}{{p}} \\ $$$$\Rightarrow{tg}\frac{{A}}{\mathrm{2}}={tg}\frac{{B}}{\mathrm{2}}+{tg}\frac{{C}}{\mathrm{2}}+\frac{{S}}{{p}^{\mathrm{2}} } \\ $$$${tg}\frac{{A}}{\mathrm{2}}.{tg}\frac{{B}}{\mathrm{2}}.{tg}\frac{{C}}{\mathrm{2}}=\frac{{S}}{{p}^{\mathrm{2}} } \\ $$$$\Rightarrow{tg}\frac{{A}}{\mathrm{2}}={tg}\frac{{B}}{\mathrm{2}}+{tg}\frac{{C}}{\mathrm{2}}+{tg}\frac{{A}}{\mathrm{2}}.{tg}\frac{{B}}{\mathrm{2}}.{tg}\frac{{C}}{\mathrm{2}} \\ $$$$\Rightarrow{tg}\frac{{A}}{\mathrm{2}}\left(\mathrm{1}−{tg}\frac{{B}}{\mathrm{2}}.{tg}\frac{{C}}{\mathrm{2}}\right)={tg}\frac{{B}}{\mathrm{2}}+{tg}\frac{{C}}{\mathrm{2}} \\ $$$$\Rightarrow{tg}\frac{{A}}{\mathrm{2}}=\frac{{tg}\frac{{B}}{\mathrm{2}}+{tg}\frac{{C}}{\mathrm{2}}}{\mathrm{1}−{tg}\frac{{B}}{\mathrm{2}}.{tg}\frac{{C}}{\mathrm{2}}}={tg}\left(\frac{{B}}{\mathrm{2}}+\frac{{C}}{\mathrm{2}}\right)\Rightarrow \\ $$$$\frac{{A}}{\mathrm{2}}=\frac{{B}+{C}}{\mathrm{2}}\Rightarrow{A}={B}+{C}=\mathrm{180}−{A}\Rightarrow{A}=\mathrm{90}^{\bullet} \:.\blacksquare \\ $$
Commented by Tinkutara last updated on 20/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 20/Jun/17
r_1 =(Δ/(s−a)) r_2 +r_3 +r=(Δ/(s−b))+(Δ/(s−c))+(Δ/s) when above r_1 =r_2 +r_3 +r (1/(s−a))=(1/(s−b))+(1/(s−c))+(1/s) (1/(s−a))−(1/s)=(1/(s−b))+(1/(s−c)) (a/(s(s−a)))=((2s−(b+c))/((s−b)(s−c))) (a/(s(s−a)))=(a/((s−b)(s−c))) or (s−b)(s−c)=s(s−a) bc−s(b+c)=−as bc=s(b+c−a) bc=(((a+b+c))/2)(b+c−a) 2bc=(b+c+a)(b+c−a) 2bc=b^2 +bc−ab+bc+c^2 −ac+ab +ac−a^2 ⇒ a^2 =b^2 +c^2 .
$${r}_{\mathrm{1}} =\frac{\Delta}{{s}−{a}}\: \\ $$$${r}_{\mathrm{2}} +{r}_{\mathrm{3}} +{r}=\frac{\Delta}{{s}−{b}}+\frac{\Delta}{{s}−{c}}+\frac{\Delta}{{s}} \\ $$$${when}\:{above}\:{r}_{\mathrm{1}} ={r}_{\mathrm{2}} +{r}_{\mathrm{3}} +{r} \\ $$$$\:\frac{\mathrm{1}}{{s}−{a}}=\frac{\mathrm{1}}{{s}−{b}}+\frac{\mathrm{1}}{{s}−{c}}+\frac{\mathrm{1}}{{s}} \\ $$$$\:\frac{\mathrm{1}}{{s}−{a}}−\frac{\mathrm{1}}{{s}}=\frac{\mathrm{1}}{{s}−{b}}+\frac{\mathrm{1}}{{s}−{c}} \\ $$$$\frac{{a}}{{s}\left({s}−{a}\right)}=\frac{\mathrm{2}{s}−\left({b}+{c}\right)}{\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$\:\frac{{a}}{{s}\left({s}−{a}\right)}=\frac{{a}}{\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${or}\:\:\:\:\left({s}−{b}\right)\left({s}−{c}\right)={s}\left({s}−{a}\right) \\ $$$$\:\:\:\:\:\:\:{bc}−{s}\left({b}+{c}\right)=−{as} \\ $$$$\:\:\:\:\:\:\:{bc}={s}\left({b}+{c}−{a}\right) \\ $$$$\:\:\:\:\:\:\:{bc}=\frac{\left({a}+{b}+{c}\right)}{\mathrm{2}}\left({b}+{c}−{a}\right) \\ $$$$\:\:\:\:\:\:\mathrm{2}{bc}=\left({b}+{c}+{a}\right)\left({b}+{c}−{a}\right) \\ $$$$\:\mathrm{2}{bc}={b}^{\mathrm{2}} +{bc}−{ab}+{bc}+{c}^{\mathrm{2}} −{ac}+{ab} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{ac}−{a}^{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:\:\boldsymbol{{a}}^{\mathrm{2}} =\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 20/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by mrW1 last updated on 20/Jun/17
with r_1 = r_2 + r_3 + r we have (Δ/(s−a))=(Δ/(s−b))+(Δ/(s−c))+(Δ/s) (1/(s−a))−(1/s)=(1/(s−b))+(1/(s−c)) (a/(s(s−a)))=((2s−b−c)/((s−b)(s−c))) (a/(s(s−a)))=(a/((s−b)(s−c))) ⇒s(s−a)=(s−b)(s−c) sa−s(b+c)+bc=0 s(a−b−c)+bc=0 (a+b+c)(a−b−c)+2bc=0 a^2 −(b+c)^2 +2bc=0 a^2 −b−c^2 −2bc+2bc=0 ⇒a^2 =b+c^2 ⇒triangle is right angled!
$$\mathrm{with}\:{r}_{\mathrm{1}} \:=\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:+\:{r}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\Delta}{\mathrm{s}−\mathrm{a}}=\frac{\Delta}{\mathrm{s}−\mathrm{b}}+\frac{\Delta}{\mathrm{s}−\mathrm{c}}+\frac{\Delta}{\mathrm{s}} \\ $$$$\frac{\mathrm{1}}{\mathrm{s}−\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{s}}=\frac{\mathrm{1}}{\mathrm{s}−\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{s}−\mathrm{c}} \\ $$$$\frac{\mathrm{a}}{\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)}=\frac{\mathrm{2s}−\mathrm{b}−\mathrm{c}}{\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$\frac{\mathrm{a}}{\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)}=\frac{\mathrm{a}}{\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$\Rightarrow\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)=\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right) \\ $$$$\mathrm{sa}−\mathrm{s}\left(\mathrm{b}+\mathrm{c}\right)+\mathrm{bc}=\mathrm{0} \\ $$$$\mathrm{s}\left(\mathrm{a}−\mathrm{b}−\mathrm{c}\right)+\mathrm{bc}=\mathrm{0} \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}−\mathrm{b}−\mathrm{c}\right)+\mathrm{2bc}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} −\left(\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} +\mathrm{2bc}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}−\mathrm{c}^{\mathrm{2}} −\mathrm{2bc}+\mathrm{2bc}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} =\mathrm{b}+\mathrm{c}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{triangle}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled}! \\ $$
Commented by Tinkutara last updated on 20/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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