Question

ABEC IS A CIRCUM CIRCLE WITH CENTRE O OF TRIANGLE ABC. AB IS THE HEIGHT OF THE TRIANGLE ABC AND IS THE DIAMETER OF THE CIRCUMCIRCLE PROVE THAT ANGLE BAD IS EQUAL TO ANGLE CAE

Answer 1

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.

Answer 2

Answer:

here is the answer

Step-by-step explanation:

To prove : angle OBC +angle BAC + = 90

Construction : Draw OD BC =

Proof : In Δ OBD and Δ OCD, we have

OB = OC [Radii of the same circle]

angle ODB =   angle ODC = 90

[By construction]

OD = OD [Common side]

` Δ OBD is congruent to Δ OCD [By RHS congruence]

angle 3 = angle 4 [C.P.C.T.]

angle 1 = angle 2 [C.P.C.T.]

angle BOC = 2angle BAC

& 2 3 + = 2 angle BAC

[a angle BOC = + + 3 4 + = + + 3 3 + = 2 3 + ]

& +3 = +BAC

+4 = +BAC

+ + 3 2 + = + + BAC + 2

[Adding +2 both side]

90 = angle BAC + 1

[a + + 1 2 = and + + 3 4 = Also, + + 3 4 90 + = c]

& + + OBC A + = 90c

Hence proved.