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Ex 7.3, 2 - AD is an altitude of an isosceles triangle ABC - Ex 7.3

Ex 7.3, 2 - Chapter 7 Class 9 Triangles - Part 2

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Ex 7.3,2 AD is an altitude of an isosceles triangle ABC in which AB = AC . Show that (i) AD bisects BC , (ii) AD bisects āˆ š“. Given: āˆ† ABC is an isosceles triangle, So, AB = AC Also, AD is the altitude So, āˆ š“DC = āˆ š“DB = 90āˆ˜ To prove: (i) BD = CD & (ii) āˆ šµš“š· = āˆ š¶š“š· Proof In āˆ†ADB and āˆ†ADC āˆ š“DC = āˆ š“DB = 90Ā° AB = AC AD = AD āˆ“ āˆ† ADB ā‰… āˆ† ADC Hence, by CPCT ā‡’ BD = DC and āˆ šµš“š¶ = āˆ š·š“š¶ 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.