Help with Triangle problems please!

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.

Easy Math Editor

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Comments

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

Mehul Arora - 5 years, 10 months ago

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 $\angle ADB > \angle BAD\Rightarrow AB>BD$

Anik Mandal - 5 years, 10 months ago

Not that poor in geometry.. xD

Anik Mandal - 5 years, 10 months ago

@Anik Mandal – Haha, I know xD

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

Mehul Arora - 5 years, 10 months ago

@Anik Mandal Thanks! ^_^ I was not really able to figure it out. It was an easy problem though. Thanks so much :)

Mehul Arora - 5 years, 10 months ago

@Mehul Arora – Welcome bro. Did you get the first two?

Anik Mandal - 5 years, 10 months ago

@Anik Mandal – Nah, Not really :/ :/

Mehul Arora - 5 years, 10 months ago

For the 1st problem.

User 123 - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$