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

part 2 - Ex 7.3, 4 - Ex 7.3 - Serial order wise - Chapter 7 Class 9 Triangles

Remove Ads

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.