If ad=bd=cd then prove that angle bac is a right angle
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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,
∠ ABC = 90°
So,
∆ ABC is a right angle triangle . ( Hence proved )
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,
∠ ABC = 90°
So,
∆ ABC is a right angle triangle . ( Hence proved )
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