A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA 2 +EB 2 +EC 2 +ED 2 =4R 2

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pavaneswari · Answer

you say A,B,C,D are 4 points on cirrcle with radius R and AC perpendicular to BD from these we say AC and BD are meets at center of the circle.so EA=EB=EC=ED=R that's mean EA2+EB2+EC2+ED2=4R2

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A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA 2 +EB 2 +EC 2 +ED 2 =4R 2

A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA²+EB²+EC²+ED²=4R²

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A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA 2 +EB 2 +EC 2 +ED 2 =4R 2

A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA2+EB2+EC2+ED2=4R2

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Magical Mathematics[Interesting Approach]

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A,B,C,D are four points on a circle with radius R such that AC is perpendicular to BD and meets BD at E. prove that:EA²+EB²+EC²+ED²=4R²