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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 5 In Fig. 6.60, AD is a median of a triangle ABC and AM โŠฅ BC. Prove that : AC2 = AB2 + BC2 โ€“ 2 BC . DM + (๐ต๐ถ/2)^2. Given: ABC is a triangle AD is a Median of โˆ†ABC โˆด BD = CD = 1/2 BC Also, AM โŠฅ BC To Prove: (i) AC2 = AB2 + BC2 โ€“2 BC . DM + (๐ต๐ถ/2)^2. Proof : In Right โˆ†AMD Applying Pythagoras Theorem, AD2 = AM2 + DM2 In Right โˆ†AMC Applying Pythagoras Theorem, AC2 = AM2 + CM2 Thus, our equations are AD2 = AM2 + DM2 โ€ฆ(2) AC2 = AM2 + CM2 โ€ฆ(3) Subtracting (3) โ€“ (2) AC2 โ€“ AD2 = AM2 + CM2 โ€“ (AM2 + DM2) AC2 โ€“ AD2 = AM2 + CM2 โ€“ AM2 โ€“ DM2 AC2 โ€“ AD2 = CM2 โ€“ DM2 AC2 = AD2 + CM2 โ€“ DM2 Putting CM = CD + DM AC2 = AD2 + (CD + DM)2 โˆ’ DM2 AC2 = AD2 + CD2 + DM2 + 2CD.DM โˆ’ DM2 AC2 = AD2 + CD2 + 2CD.DM Putting CD = ๐Ÿ/๐Ÿ BC Form equation (1) AC2 = AD2 + (๐‘ฉ๐‘ช/๐Ÿ)^2+ 2(๐ต๐ถ/2) (DM) AC2 = AD2 + BC.DM + (๐ต๐ถ/2)^2 Hence Proved. Ex 6.6, 5 In Fig. 6.59, ABC is a triangle in which โˆ  ABC < 90ยฐ and AD โŠฅ BC.Prove that (ii) AB2 = AD2 โ€“ BC.DM + [๐ต๐ถ/2]^2. In Right โˆ†AMB Applying Pythagoras Theorem, AB2 = AM2 + BM2 In Right โˆ†AMD Applying Pythagoras Theorem, AD2 = AM2 + DM2 Thus, our equations are AB2 = AM2 + BM2 โ€ฆ(4) AD2 = AM2 + DM2 โ€ฆ(5) Subtracting (4) โ€“ (5) AB2 โ€“ AD2 = AM2 + BM2 โ€“ (AM2 + DM2) AB2 โ€“ AD2 = AM2 + BM2 โ€“ AM2 โ€“ DM2 AB2 โ€“ AD2 = BM2 โ€“ DM2 AB2 = AD2 + BM2 โ€“ DM2 Putting BM = BD โ€“ DM AB2 = AD2 + (BD โ€“ DM)2 โ€“ DM2 AB2 = AD2 โˆ’ DM2 + (BD โˆ’ DM)2 AB2 = AD2 โˆ’ DM2 + BD2 + DM2 โˆ’ 2BD.DM AB2 = AD2 + BD2 โˆ’ 2BD.DM Putting BD = 1/2 BC from equation (1) AB2 = AD2 + (๐‘ฉ๐‘ช/๐Ÿ)^2 โˆ’ 2 (๐‘ฉ๐‘ช/๐Ÿ).DM AB2 = AD2 โˆ’ BC.DM + (๐ต๐ถ/2)^2 Hence Proved Ex 6.6, 5 In Fig. 6.59, ABC is a triangle in which โˆ  ABC < 90ยฐ and AD โŠฅ BC. Prove that (iii) AC2 = AB2 = 2 AD2 + 1/2 BC2 . Results of (i) and (ii) part are AC2 = AD2 + BC.DM + (๐ต๐ถ/2)^2 โ€ฆ(6) AB2 = AD2 โˆ’ BC.DM + (๐ต๐ถ/2)^2 โ€ฆ(7) Adding equations (6) & (7) AC2 + AB2 = AD2 + BC.DM + (๐ต๐ถ/2)^2 + AD2 โˆ’ BC.DM + (๐ต๐ถ/2)^2 AC2 + AB2 = 2AD2 + (๐ต๐ถ/2)^2 + (๐ต๐ถ/2)^2 AC2 + AB2 = 2AD2 + (๐ต๐ถ^2)/4 + (๐ต๐ถ^2)/4 AC2 + AB2 = 2AD2 + 2((๐ต๐ถ^2)/4) AC2 + AB2 = 2AD2 + 1/2 ๐ต๐ถ^2 Hence Proved

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