Check sibling questions

Ex 7.3, 4 - BE and CF are two equal altitudes of triangle ABC - RHS

Ex 7.3, 4 - Chapter 7 Class 9 Triangles - Part 2

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 7.3,4 BE and CF are two equal altitudes of a triangle ABC . Using RHS congruence rule , prove that the triangle ABC is isosceles . Given: Given BE is a altitude, So, ∠𝐴EB = ∠CEB= 90∘ Also, CF is a altitude, So, ∠𝐴FC = ∠BFC= 90∘ Also, BE = CF To prove: Δ ABC is isoceles Proof: In ∆BCF and ∆CBE ∠BFC = ∠CEB = 90∘ BC = CB FC = EB ∆ BCF ≅ ∆ CBE ∴ ∠FBC= ∠ECB So, ∠ABC = ∠ACB AB = AC So, ∆ABC is an isosceles triangle

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.