Question

in triangle abc if D is a point on AC such that AD= CD = BD . then prove that triangle ABC is a right angled triangle

Answer 1

Given :- In ∆ ABC, D is the point on AC, such that AD = CD = BD

To Prove :- ∆ ABC is right angled triangle.

Proof :- In ∆ ABC :- AD = BD = CD

Hence, we know that angles opposite to equal sides are also equal.

So, angle DAB = angle DBA = let x
and, angle DBC = angle DCB = let y

Now, from angle sum property of a triangle :-

=> x + x + y + y = 180°
=> 2(x + y) = 180°
=> x + y = 90°

Therefore, angle ABC = angle DAB + angle DCB = 90°

Thus, ∆ ABC is a right - angled triangle.

Q.E.D.





Hope this helps.....

Answer 2

Answer:

Given :- In ∆ ABC, D is the point on AC, such that AD = CD = BD

To Prove :- ∆ ABC is right angled triangle.

Proof :- In ∆ ABC :- AD = BD = CD

Hence, we know that angles opposite to equal sides are also equal.

So, angle DAB = angle DBA = let x

and, angle DBC = angle DCB = let y

Now, from angle sum property of a triangle :-

=> x + x + y + y = 180°

=> 2(x + y) = 180°

=> x + y = 90°

Therefore, angle ABC = angle DAB + angle DCB = 90°

Thus, ∆ ABC is a right - angled