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प्रश्न
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) `"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
(ii) `"AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2`
(iii) `"AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2`
उत्तर
(i) Applying Pythagoras theorem in ΔAMD, we obtain
AM2 + MD2 = AD2 … (1)
Applying Pythagoras theorem in ΔAMC, we obtain
AM2 + MC2 = AC2
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)]
Using the result, DC = `"BC"/2`, we obtain
`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`
`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`
(ii) Applying Pythagoras theorem in ΔABM, we obtain
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD)2
= AD2 − DM2 + BD2 + MD2 − 2BD × MD
= AD2 + BD2 − 2BD × MD
= `"AD"^2+(("BC")/2)^2 - 2(("BC")/2) xx "MD"`
= `"AD"^2 + ("BC"/2)^2 - "BC" xx "MD"`
(ii) Applying Pythagoras theorem in ΔABM, we obtain
AM2 + MB2 = AB2 … (1)
Applying Pythagoras theorem in ΔAMC, we obtain
AM2 + MC2 = AC2 … (2)
Adding equations (1) and (2), we obtain
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2
2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC)
= AB2 + AC2
= `2("AM"^2 + "MD"^2) + (("BC")/2)^2 + (("BC")/2)^2 + 2"MD" ((-"BC")/2 + ("BC")/2) = "AB"^2 + "AC"^2`
`"2AD"^2 + ("BC"^2)/2 = "AB"^2 + "AC"^2`
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