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?ABC is an isosceles right angled triangle having ?C = 90°. If D is any point on AB, then AD2+ BD2 is equal to
CD2
2CD2
3CD2
4CD2
?ABC is a right angle triangle. In which ∠C=90° And D is a point on AB such that D is perpendicular on AB. Let AC = BC = a ∴AB^2=AC^2+BC^2=a^2+a^2 AB=a^2 ∴BD=AD=a√2/2=a/√2 Now In ?ACD =AC^2=CD^2+AD^2 a^2=CD^2+a^2/2 a^2−a^2/2=CD^2 a^2/2=CD^2 CD^2=a^2/2 2CD^2=a^2 and AD^2+BD^2=(a/√2)^2+(a/√2)^2 =a/√2+a/√2=a^2 ∴2CD^2=AD^2+BD^2
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error
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