आपकी दृष्टि में मास्टर प्रीतम चंद का व्यवहार छात्रों के साथ उचित है या अनुचित तर्कपूर्ण उत्तर दीजिए
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Let AX be a line perpendicular to BC and as perpendicular from a vertex common to equal side is also a median therefore, BX=CX=
2
1
(BC)
Now, By Pythagoras theorem in ΔADX and ΔACX,
AD
2
=AX
2
+XD
2
and AC
2
=AX
2
+CX
2
⇒AD
2
−AC
2
=XD
2
−CX
2
=(BD−BX)
2
−CX
2
=BD
2
+BX
2
−2BD.BX−CX
2
(BX=CX)
=BD(BD−2BX)
=BD(BD−BC)
⇒AD
2
−AC
2
=BD⋅CD
∴AD
2
=AC
2
+BD⋅CD
solution
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