# 1. If the bisector of an angle of a triangle also bisect the opp. side , prove that the triangle is isos . triangle ?2. If triangle ABC is right angle at B. such that angle BCA=2 angle BAC . show that hypt.AC =2BC  ?3. bisector of angle B and angle C of isos. triangle ABC with AB=AC intersect each other at O. Show that external angle adjacent to angle ABC = angle BOC 4. In a rt. triangle prove that the line segment joining the mid point of the hypt to the opp. vertex is half of the hypt ?5. line segment joining the midpoint M&N of prallel sides AB and DC . respectivelly of a trap. ABCD is perpendicular to both sides AB and DC . prove that AD = BC ?

Hi Beamy!

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).

Q 1.
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 = ∠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.

Hence, proved

Q 2.
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°

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]

Hence, proved