Question 19 (OR 2 nd Question)

In ∆ABC, if AD is the median, then show that AB 2 + AC 2 = 2(AD 2 + BD 2 )

In ABC, if AD is the median, show that AB^2 + AC^2 = 2(AD^2 + BD^2)

Question 19 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 2
Question 19 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 3 Question 19 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 4

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Transcript

Question 19 (OR 2nd Question) In ∆ABC, if AD is the median, then show that AB2 + AC2 = 2(AD2 + BD2) Given: AD is a Median of ∆ABC ∴ BD = CD Also, AE ⊥ BC To Prove: AB2 + AC2 = 2(AD2 + BD2) Proof : In Right ∆ABE Applying Pythagoras Theorem, AB2 = AE2 + BE2 In Right ∆ACE …(2) Applying Pythagoras Theorem, AC2 = AE2 + CE2 …(3) …(3) By Pythagoras theorem in Δ AED AD2 = AE2 + ED2 AE2 = AD2 – ED2 By Pythagoras theorem in Δ AED AD2 = AE2 + ED2 AE2 = AD2 – ED2 AB2 + AC2 = 2AE2 + BE2 + CE2 AB2 + AC2 = 2(AD2 – ED2) + BE2 + CE2 AB2 + AC2 = 2AD2 – 2ED2 + BE2 + CE2 Now, writing BE = BD – ED CE = CD + ED AB2 + AC2 = 2AD2 – 2ED2 + (BD – ED)2 + (CD + ED)2 AB2 + AC2 = 2AD2 – 2ED2 + BD2 + ED2 – 2BD ED + CD2 + ED2 + 2 CD ED AB2 + AC2 = 2AD2 – 2ED2 + ED2 + ED2 + BD2 + CD2 + 2 CD ED – 2BD ED AB2 + AC2 = 2AD2 + BD2 + CD2 + 2 ED (CD – BD) From (1), CD = BD AB2 + AC2 = 2AD2 + BD2 + BD2 + 2 ED (BD – BD) AB2 + AC2 = 2AD2 + 2BD2 + 2 ED × 0 AB2 + AC2 = 2AD2 + 2BD2 AB2 + AC2 = 2(AD2 + 2BD2) Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.