In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC2 = AD2 + BC.DM + 2 ($$\dfrac{BC}{2}$$) 2

(ii) AB2 = AD2 – BC.DM + 2 ($$\dfrac{BC}{2}$$) 2

(iii) AC2 + AB2 = 2 AD2 + $$\dfrac{1}{2}$$ BC2

Asked by Pragya Singh | 1 year ago |  45

##### Solution :-

(i) By applying Pythagoras Theorem in ∆AMD, we get,

AM2 + MD2 = AD2 ………………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,

AM2 + MC2 = AC2

AM2 + (MD + DC) 2 = AC2

(AM2 + MD2 ) + DC2 + 2MD.DC = AC2

From equation(i), we get,

AD2 + DC2 + 2MD.DC = AC2

Since, DC=BC/2, thus, we get,

AD+ (BC/2) 2 + 2MD.(BC/2) 2 = AC2

AD+ (BC/2) 2 + 2MD × BC = AC2

Hence, proved.

(ii) By applying Pythagoras Theorem in ∆ABM, we get;

AB2 = AM2 + MB2

= (AD2 − DM2) + MB2

= (AD2 − DM2) + (BD − MD) 2

= AD2 − DM2 + BD2 + MD2 − 2BD × MD

= AD2 + BD2 − 2BD × MD

= AD+ ($$\dfrac{BC}{2}$$)– 2($$\dfrac{BC}{2}$$) MD

= AD+ ($$\dfrac{BC}{2}$$)– BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in ∆ABM, we get,

AM2 + MB2 = AB2 ………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

AM2 + MC2 = AC2 ……..… (ii)

Adding both the equations (i) and (ii), we get,

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(AM2+ MD2) + ($$\dfrac{BC}{2}$$) 2 + ($$\dfrac{BC}{2}$$) 2 + 2MD

($$\dfrac{-BC}{2}$$ + $$\dfrac{BC}{2}$$) 2 = AB2 + AC2

2AD+ $$\dfrac{BC^2}{2}$$ = AB2 + AC2

Answered by Abhisek | 1 year ago

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