Q1. ABC is a triangle with Angle B > 2 Angle C D is a point on BC such that AD bisects Angle BAC and AB=CD. Prove that Angle BAC=
Q2. AD, BE and CF are medians of a triangle ABC. Prove that 2(AD+BE+CF)<3(AB+BC+CA)<4(AD+BE+CF)
Q3. In Triangle ABC, AD is the bisector of Angle BAC Prove that AB>BD.
Easy Math Editor
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@Chew-Seong Cheong Sir, @Nihar Mahajan @Anik Mandal @Archit Boobna
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If you draw the diagram for the 3rd question you get
∠ADB=∠2A+∠C
whereas ∠BAD=∠2A
since ∠ADB>∠BAD⇒AB>BD
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Not that poor in geometry.. xD
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Neither am I. Idk why I was unable to figure this out :/
@Anik Mandal Thanks! ^_^ I was not really able to figure it out. It was an easy problem though. Thanks so much :)
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For the 1st problem.