In-AB-C-B-90-o-BB-CC-BB-and-CC-are-medians-m-c-m-a-note-CC-m-c-AA-m-a-

Question Number 207980 by mnjuly1970 last updated on 01/Jun/24

I n A \overset{Δ}{B C} : B = 90^{o}

{BB}^{'} ⊥ {CC}^{'} ({BB}^{'} a n d {CC}^{'} a r e m e d i a n s)

\Rightarrow \frac{m_{c}}{m_{a}} = ?

n o t e : ∣ {CC}^{'} ∣ = m_{c}, ∣ {AA}^{'} ∣= m_{a}

Answered by mr W last updated on 02/Jun/24

Commented by mr W last updated on 02/Jun/24

B^{'} G = \frac{m_{b}}{3} = \frac{\sqrt{2 (a^{2} + c^{2}) - b^{2}}}{6}

C G = \frac{2 m_{c}}{3} = \frac{\sqrt{2 (a^{2} + b^{2}) - c^{2}}}{3}

B^{'} C = \frac{b}{2}

{(\frac{\sqrt{2 (a^{2} + c^{2}) - b^{2}}}{6})}^{2} + {(\frac{\sqrt{2 (a^{2} + b^{2}) - c^{2}}}{3})}^{2} = {(\frac{b}{2})}^{2}

\frac{2 (a^{2} + c^{2}) - b^{2}}{36} + \frac{2 (a^{2} + b^{2}) - c^{2}}{9} = \frac{b^{2}}{4}

\Rightarrow 5 a^{2} = b^{2} + c^{2}

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

\Rightarrow 2 a^{2} = c^{2}

\frac{m_{c}}{m_{a}} = \sqrt{\frac{2 (a^{2} + b^{2}) - c^{2}}{2 (b^{2} + c^{2}) - a^{2}}}

= \sqrt{\frac{2 (2 a^{2} + 2 a^{2}) - 2 a^{2}}{2 (a^{2} + 4 a^{2}) - a^{2}}} = \sqrt{\frac{2}{3}} ✓

Commented by efronzo1 last updated on 02/Jun/24

AC is \underset{⏟}{}

Commented by mr W last updated on 02/Jun/24

y e s, s i n c e ∠ B = 90 °

Commented by mr W last updated on 02/Jun/24

Commented by mnjuly1970 last updated on 02/Jun/24

g r a t e f u l m a s t e r

v e r y n i c e a s a l w a y s . \underset{―}{\underset{⏟}{⋚}}

Commented by Tawa11 last updated on 21/Jun/24

Weldone sir

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