In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that : (i) AC^2 = AD^2

Question 5

In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i)$ AC^2 = AD^2 + BC . DM + (\dfrac{BC}{2})^2$

(ii) $AB^2 = AD^2 – BC . DM + (\dfrac{BC}{2})^2$

(iii) $AC^2 + AB^2 = 2 AD^2 + \dfrac{1}{2} BC^2$

Solution:

(i) $AC^2 = AD^2 + BC . DM + (\dfrac{BC}{2})^2$

By applying Pythagoras Theorem in ∆AMD, we get,

$AM^2 + MD^2 = AD^2 $………………. (i)

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

$AM^2 + MC^2 = AC^2$

$AM^2 + (MD + DC)^2 = AC^2$

$(AM^2 + MD^2 ) + DC^2 + 2MD.DC = AC^2$

From equation(i), we get,

$AD^2 + DC^2 + 2MD.DC = AC^2$

Since, DC=$\dfrac{BC}{2}$, thus, we get,

$AD^2 + (\dfrac{BC}{2})^2 + 2MD.(\dfrac{BC}{2})^2 = AC^2$

$AD^2 + (\dfrac{BC}{2})^2 + 2MD × BC = AC^2$

Hence, proved.

(ii) $AB^2 = AD^2 – BC . DM + (\dfrac{BC}{2})^2$

By applying Pythagoras Theorem in ∆ABM, we get

$AB^2 = AM^2 + MB^2$

= $(AD^2 − DM^2) + MB^2$

= $(AD^2 − DM^2) + (BD − MD)^2$

= $AD^2 − DM^2 + BD^2 + MD^2 − 2BD × MD$

= $AD^2 + BD^2 − 2BD × MD$

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

= $AD^2 + (\dfrac{BC}{2})^2 – BC MD$

Hence, proved.

(iii) $AC^2 + AB^2 = 2 AD^2 + \dfrac{1}{2} BC^2$

By applying Pythagoras Theorem in ∆ABM, we get,

$AM^2 + MB^2 = AB^2$ ………………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

$AM^2 + MC^2 = AC^2$ …………………..… (ii)

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

$2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$

$2AM^2 + (BD − DM)^2 + (MD + DC)^2 = AB^2 + AC^2$

$2AM^2+BD^2 + DM^2 − 2BD.DM + MD^2 + DC^2 + 2MD.DC = AB^2 + AC^2$

$2AM^2 + 2MD^2 + BD^2 + DC^2 + 2MD (− BD + DC) = AB^2 + AC^2$

$2(AM^2+ MD^2) + (\dfrac{BC}{2})^2 + (\dfrac{BC}{2})^2 + 2MD (\dfrac{-BC}{2} + \dfrac{BC}{2})^2 = AB2 + AC^2$

$2AD^2 + \dfrac{BC^2}{2} = AB^2 + AC^2$

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