आकृति में AD त्रिभुज ABC की एक माध्यिका है तथा AM⊥BC है। सिद्ध कीजिए कि
(i) $AC squared equals AD squared plus BC. DM plus open parentheses BC over 2 close parentheses squared$
(ii) $AB squared space equals space AD squared minus BC. space DM plus open parentheses BC over 2 close parentheses squared$
(iii) $AC squared plus AB squared space equals space 2 AD squared plus 1 half BC squared$

Solution

(i) समकोण ΔACM में,
AC² = AM² + MC²

equals space space space space space AM squared plus left parenthesis MD plus DC right parenthesis squared equals space space space space space AM squared plus MD squared plus DC squared plus 2 MD. DC equals space space space space space AD squared plus open parentheses 1 half BC close parentheses squared plus 2 DM.1 half BC equals space space space space space AD squared plus 1 fourth BC squared plus BC. DM space space space space space space space space space space... left parenthesis straight i right parenthesis

(ii) समकोण ΔABM में,
AB² = AM² + BM²
= AM² + (BD - MD)²= AM² + BD² + MD² - 2BD.MD
= (AM² + MD²) + BD² - 2BD.MD
=

space space space space space AD squared plus open parentheses 1 half BC close parentheses squared minus 2 cross times 1 half BC. DM

equals space space space space space AD squared plus 1 fourth BC squared minus BC. DM space space space space... left parenthesis ii right parenthesis

(iii) समीकरण (i) और (ii) को जोड़ने पर

AC squared plus AB squared equals 2 AD squared plus 1 half BC squared

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