# 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=$$72^{\circ}$$

Q3. In Triangle ABC, AD is the bisector of Angle BAC Prove that AB>BD.

Note by Mehul Arora
2 years, 12 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

- 2 years, 12 months ago

For the 1st problem.

- 2 years, 12 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 $$\angle ADB > \angle BAD\Rightarrow AB>BD$$

- 2 years, 12 months ago

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

- 2 years, 12 months ago

Welcome bro. Did you get the first two?

- 2 years, 12 months ago

Nah, Not really :/ :/

- 2 years, 12 months ago

Not that poor in geometry.. xD

- 2 years, 12 months ago

Haha, I know xD

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

- 2 years, 12 months ago