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in Triangles by (34.5k points)
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In ΔABC, D is the mid point of BC. Then AB2 + AC2 =

(A) AD2 + BD2

(B) 2AD2 + 2BD2

(C) BD2 + DC2

(D) 2AC2 + 2CD2

2 Answers

+1 vote
by (57.0k points)
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Best answer

Correct option is (B) \(2AD^2 + 2BD^2\)

\(\because\) D is mid-point of BC.

\(\therefore\) BD = CD

Construction :-

Draw \(AE\bot BC\)

In right \(\triangle AED,\)

\(AD^2=AE^2+DE^2\) _______________(1)

In right \(\triangle AEB,\)

\(AB^2=BE^2+AE^2\) _______________(2)

In right \(\triangle AEC,\)

\(AC^2=AE^2+CE^2\) _______________(3)

Now,  \(AB^2+AC^2\) \(=(BE^2+AE^2)+(AE^2+CE^2)\)   (From (2) & (3))

\(=2AE^2+BE^2+CE^2\)

\(=2(AD^2-DE)^2+BE^2+CE^2\)   (From (1))

\(=2AD^2+BE^2+CE^2-2DE^2\)

\(=2AD^2+(BD-DE)^2+(CD+DE)^2-2DE^2\) \((\because BE=BD-DE\;\&\;CE=CD+DE)\)

\(=2AD^2+BD^2+DE^2-2BD.DE\) \(+CD^2+DE^2+2CD.DE-2DE^2\)

\(=2AD^2+BD^2-2BD.DE+BD^2+2BD.DE\) \((\because CD=BD)\)

\(=2AD^2+2BD^2\)

\(=2(AD^2+BD^2)\)

Hence, \(AB^2+AC^2=2AD^2+2BD^2\)

+1 vote
by (35.6k points)

Correct option is: (B) 2AD2 + 2BD2

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