1 If the bisector of an angle of a triangle also bisect the opp side , prove that - Maths - Triangles

Question

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 ?

Mayank Bhardwaj · Accepted Answer

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 âˆ†ABC, let AD be the bisector of ∠A and BD = CD It is required to prove âˆ†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) CE||AD ∴ ∠BAD = ∠AEC (Corresponding angles) (2) And ∠CAD = ∠ACE (Alternate interior angles) (3) From (1), (2) and (3) ∠ACE = ∠AEC In âˆ†ACE, ∠ACE = ∠AEC ∴ AE = OAC (Sides opposite to equal angles) (4) In âˆ†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 âˆ†ABC, AB = AC ∴ âˆ†ABC is an isosceles triangle Hence, proved Q 2 Consider the âˆ†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 âˆ†ACD, ∠CAD = ∠ACD = ∠ADC = 60° So, âˆ†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 Hope! This will help you Cheers!!!