We are happy to see the interest you are showing towards this section! Thank you for your response. However, due to a paucity of time we will not be able to immediately answer all your queries. We are providing the answer to questions (1) and (2).
Consider the ∆ABC, let AD be the bisector of ∠A and BD = CD. It is required to prove ∆ABC is an isosceles triangle i.e. AB = AC. For this draw a line from C parallel AD and extend BA. Let they meet at E.
It is given that
∠BAD = ∠CAD ... (1)
∴ ∠BAD = ∠AEC (Corresponding angles) ... (2)
And ∠CAD = ∠ACE (Alternate interior angles) ... (3)
From (1), (2) and (3)
∠ACE = ∠AEC
In ∆ACE, ∠ACE = ∠AEC
∴ AE = OAC (Sides opposite to equal angles) ... (4)
In ∆BEC, AD||CE and D is the mid-point of BC using converse of mid-point theorem A is the mid-point of BE.
∴ AB = AE
⇒ AB = AC [Using (4)]
In ∆ABC, AB = AC
∴ ∆ABC is an isosceles triangle.
Consider the ∆ABC right angle at B and ∠BCA = BAC extend CB to D such that BD = BC. Join AD
Using angle sum property of triangle ABC
∠ABC + ∠BAC + ∠ACB = 180°
⇒ 90° + ∠BAC + 2∠BAC = 180°
⇒ 3∠BAC = 90°
⇒∠BAC = 30°
∴ ∠ACB = 2 × 30° = 60°
Similarly, it can be proved that ∠ADB = 60° and ∠BAD = 30°
∴ ∠CAD = 30° + 30° = 60°
In ∆ACD, ∠CAD = ∠ACD = ∠ADC = 60°
So, ∆ABC is an equilateral triangle.
∴ AC = CD = AD
It is know that the perpendicular drawn to a side from opposite vertex bisects the side
∴ CD = 2BC
⇒ AC = 2BC [CD = AC]
Hope! This will help you.