Question

abec is circumcircle (with centr o )of a triangle abc ,ad is the height of the triangle abc and ae is a diameter of this circumfernce .prove that angle bad=angle cae​

Answer 1

Answer:

Step-by-step explanation:

Join CE.

Angle ACE =90°(Angle in a semicircle)

Angle AEC = Angle ABC (Angles in the same segment)

In Triangle AEC and Triangle ABD-

 Angle ACE = Angle ADE

 Angle AEC = Angle ABE

Then by Angle Sum Property,

  Angle BAE = Angle CAE

Answer 2

Answer:

Step-by-step explanation:

Let a be the inscribed angle ∠BAC, then the angle BOC is equal to the double value of a, since the center of the circle and BC make an angle which is the double of the inscribed angle.

Thus, we have:

=> ∠BOC = 2a

Then, we can conclude that:

∠BOD = ∠COD = a ( since ΔODB ≅ ΔODC)

In right triangle ΔODB we have that,

∠ODB = 90° (since OF ⊥ BC)

∠BOD = a

Hence, ∠OBD = 180- (90+a) =

= 90 - a =

= ∠OBC

Hence, we can say that our conclusion is that the equivalent process can be executed with the angles ∠BAD and ∠CAE.