Question-185924

Question Number 185924 by Mingma last updated on 29/Jan/23

Answered by HeferH last updated on 29/Jan/23

Commented by HeferH last updated on 30/Jan/23

i . \frac{D E}{b} = \frac{a}{c} \Rightarrow D E = \frac{a b}{c}

\frac{C E}{l} = \frac{b}{c} \Rightarrow C E = \frac{b l}{c}

\frac{C D}{C E} = \frac{D B}{E B} \Rightarrow \frac{b}{\frac{b l}{c}} = \frac{a}{E B} \Rightarrow E B = \frac{a l}{c}

i i . C E + E B = C B \Rightarrow

\frac{b l}{c} + \frac{a l}{c} = l

a + b = c

a^{2} + b^{2} + 2 a b = c^{2}

i i i . C E \cdot E B = A E \cdot E D \Rightarrow \frac{{a b l}^{2}}{c^{2}} = \frac{(c^{2} - a b) a b}{c^{2}}

c^{2} - l^{2} = a b

i v . a^{2} + b^{2} + 2 a b = c^{2}

a^{2} + b^{2} + 2 (c^{2} - l^{2}) = c^{2}

a^{2} + b^{2} + 2 c^{2} - 2 l^{2} = c^{2}

l^{2} = \frac{a^{2} + b^{2} + c^{2}}{2}

Answered by mr W last updated on 30/Jan/23

l^{2} = a^{2} + b^{2} + a b \dots (i)

l^{2} = b^{2} + c^{2} - b c \dots (i i)

l^{2} = a^{2} + c^{2} - a c \dots (i i i)

(i i) - (i i i) :

(a - b) c = (a - b) (a + b)

\Rightarrow c = a + b

2 \times (i) :

2 l^{2} = a^{2} + b^{2} + a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + {(a + b)}^{2}

2 l^{2} = a^{2} + b^{2} + c^{2}

\Rightarrow l^{2} = \frac{a^{2} + b^{2} + c^{2}}{2}

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