If in triangle ABC AD is median and AE perpendicular BC then prove that AB^+AC^=1/2BC^
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Answer:
Step-by-step explanation:
Draw AE perpendicular to BC.
Since angle AED = 900,
Therefore, in triangle ADE
angle ADE < 900 and angle ADB > 900
Thus, triangle ABD is obtuse angled triangle and triangle ADC is acute angled triangle.
Triangle ABD is obtuse angled at D and AE is perpendicular to BD produced,
Therefore,
AB2 = AD2 + BD2 + 2 BD x DE (i)
Triangle ACD is acute angled at D and AE is perpendicular to BD produced,
Therefore,
AC2 = AD2 + DC2 - 2 DC x DE
AC2 = AD2 + BD2 - 2 BD x DE (ii) (since CD = BD)
Adding (i) and (ii)
AB2 + AC2 = 2AD2 + 2BD2
AB2 +AC2 = 2AD2 + BD2
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