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# In Figure, AD is a median of a triangle ABC and AM $$\perp$$ BC. Prove that : $$(i) AC^2 = AD^2 + BC.DM + 2 (\frac{BC}{2} )^2$$ $$(ii) AB^2 = AD^2 – BC.DM + 2 (\frac{BC}{2} )^2$$ $$(iii) AC^2 + AB^2 = 2 AD^2 + {{1} \over {2}} BC^2$$

(i) By applying Pythagoras Theorem in ?AMD, we get,
$$AM^2 + MD^2 = AD^2$$ ………………. (i)

Again, by applying Pythagoras Theorem in $$\triangle$$ AMC, we get,
$$\Rightarrow AM^2 + MC^2 = AC^2$$
$$\Rightarrow AM^2 + (MD + DC)^ 2 = AC^2$$
$$\Rightarrow (AM^2 + MD^2 ) + DC^2 + 2MD.DC = AC^2$$

From equation(i), we get,
$$AD^2 + DC^2 + 2MD.DC = AC^2$$

Since, DC=$$\frac{BC}{2}$$ , thus, we get,
$$\Rightarrow AD^2 + (BC/2)^ 2 + 2MD.(BC/2)^ 2 = AC^2$$
$$\Rightarrow AD^2 + (BC/2)^ 2 + 2MD × BC = AC^2$$
Hence, proved.

(ii) By applying Pythagoras Theorem in -ABM, we get;
$$\Rightarrow AB^2 = AM^2 + MB^2$$
$$\Rightarrow (AD^2 - DM^2 ) + MB^2$$
$$\Rightarrow (AD^2 - DM^2 ) + (BD - MD)^2$$
$$\Rightarrow AD^2 - DM^2 + BD^2 + MD^2 - 2BD × MD$$
$$\Rightarrow AD^2 + BD^2 - 2BD × MD$$
$$\Rightarrow AD^2 + (\frac{BC}{2} )^2 – 2 \frac{BC}{2} MD$$
$$\Rightarrow AD^2 + (\frac{BC}{2} )^2 – BC.MD$$
Hence, proved.

(iii) By applying Pythagoras Theorem in $$\triangle$$ ABM, we get,
$$AM^2 + MB^2 = AB^2$$ ………………….… (i)

By applying Pythagoras Theorem in $$\triangle$$ AMC, we get,
$$\Rightarrow AM^2 + MC^2 = AC^2$$ …………………..… (ii)

Adding both the equations (i) and (ii), we get,
$$\Rightarrow 2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$$
$$\Rightarrow 2AM^2 + (BD - DM)^2 + (MD + DC)^2 = AB^2 + AC^2$$
$$\Rightarrow 2AM^2+BD^2 + DM^2 - 2BD.DM + MD^2 + DC^2 + 2MD.DC = AB^2 + AC^2$$
$$\Rightarrow 2AM^2 + 2MD^2 + BD^2 + DC62 + 2MD (- BD + DC) = AB^2 + AC^2$$
$$\Rightarrow 2(AM^2+ MD^2) + (\frac{BC}{2} )^2 + (\frac{BC}{2} )^2+ 2MD(\frac{-BC}{2} + \frac{BC}{2} )^ 2 = AB^2 + AC^2$$
$$\Rightarrow 2AD^2 + (\frac{BC}{2} )^2 = AB^2 + AC^2$$