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 ∠𝐴. Given: βˆ† ABC is an isosceles triangle, So, AB = AC Also, AD is the altitude So, ∠𝐴DC = ∠𝐴DB = 90∘ To prove: (i) BD = CD & (ii) ∠𝐡𝐴𝐷 = ∠𝐢𝐴𝐷 Proof In βˆ†ADB and βˆ†ADC ∠𝐴DC = ∠𝐴DB = 90Β° AB = AC AD = AD ∴ βˆ† ADB β‰… βˆ† ADC Hence, by CPCT β‡’ BD = DC and ∠𝐡𝐴𝐢 = ∠𝐷𝐴𝐢 Hence proved

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