Question

In triangle ABC if D is the point on AC such that AD=CD=BD then prove that triangle ABC is right angle.

Answer 1

Answer : 

Given  AD  = CD  =  BD  


We know from base angle theorem : The base angles theorem states that if the sides of a triangle are congruent (Isosceles triangle)then the angles opposite these sides are congruent. 

In ∆ ABD and BDC  

We get from base angle theorem  
∠ DAB  =  ∠ DBA   =  x                ---------------- ( 1 )
And
∠ DBC =  ∠ DCB    = y                    ---------------- ( 2 )
And as given AD  =  CD  =  BD  , 

SO,
Now from angle sum property in
  
So from equation 1 and 2 , we get

∠ DAB  = ∠ DBA  =  ∠ DBC  =  ∠ DCB   = x 

and we know by angle sum property  
 In ∆ ABC 

∠ DAB  +  ∠ DBA  +  ∠ DBC  +  ∠ DCB   =  180°

∠ DBA  +  ∠ DBA  +  ∠ DBC  +  ∠ DBC   =  180°

2 ∠ DBA  + 2 ∠ DBC     =  180°

2 ( ∠ DBA  +  ∠ DBC  )    =  180°

∠ DBA  +  ∠ DBC     =  90°
And
we can write

∠ DBA  +  ∠ DBC   = ∠ ABC

So,

Answer 2

Answer:

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