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=\(72^{\circ}\)

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.

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## Comments

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TopNewest@Chew-Seong Cheong Sir, @Nihar Mahajan @Anik Mandal @Archit Boobna

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For the 1st problem.

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If you draw the diagram for the 3rd question you get

\(\angle ADB = \angle \dfrac{A}{2} + \angle C\)

whereas \(\angle BAD = \angle \dfrac{A}{2}\)

since \(\angle ADB > \angle BAD\Rightarrow AB>BD\)

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@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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Not that poor in geometry.. xD

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Neither am I. Idk why I was unable to figure this out :/

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