Question

$\huge\bold\red{Q.1.Bisectors \: of \: angles \: A, \: B \: and \: C \: of \: a \: triangle \: ABC \: intersect \: its \: circumcircle \: at \: D, \: E \: and \: F \: respectively. \: Prove \: that \: the \: angles \: of \: the \: triangle \: DEF \: are \: 90° – (½)A, \: 90° – (½)B \: and \: 90° – (½)C.}\ \textless \ br /\ \textgreater$
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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Bisector of ∠A meets circumcircle at D

So, it means ∠BAD = ∠CAD = x say ----- [1]

Also,

Bisector of ∠B meets circumcircle at E

⇛ ∠CBE = ∠ABE = y say ----- [2]

Also,

Bisector of ∠C meets circumcircle at F.

⇛ ∠BCF = ∠ACF = z say ------- [3]

Now, In ∆ ABC

We know, Sum of all interior angles of a triangle is supplementary

⇛ ∠ABC + ∠BAC + ∠BCA = 180°

⇛ 2x + 2y + 2z = 180°

$\red{\rm \implies\:\boxed{ \tt{ \: x + y + z = 90 \degree \: }}} - - - (1)$

Now, We know,

Angle in same segments are equal.

Consider arc AF,

⟼ ∠ADF = ∠ACF = z [ Angles in same segment ]

Again, Consider arc AE,

⟼ ∠ADE = ∠ABE = y [ Angles in same segment ]

Now,

⟼ ∠EDF = ∠ADF + ∠ADE

⟼ ∠EDF = z + y

⟼ ∠EDF = 90° - x [ using equation (1) ]

$\rm \implies\:\boxed{ \tt{ \: \angle \: EDF \: = \: 90 \degree \: - \: \frac{1}{2} \angle A \: }}$

Similarly,

$\rm \implies\:\boxed{ \tt{ \: \angle \: DEF \: = \: 90 \degree \: - \: \frac{1}{2} \angle B \: }}$

and

$\rm \implies\:\boxed{ \tt{ \: \angle \: DFE \: = \: 90 \degree \: - \: \frac{1}{2} \angle C \: }}$

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Additional Information :-

1. Angle in semi-circle is right angle.

2. Equal chords subtends equal angles at the centre.

3. Equal chords are equidistant from the centre.

4. Perpendicular drawn from the centre of the circle bisects the chord.

5. Angle subtended at the centre by an arc is double the angle subtended on the circumference by the same arc.

6. Sum of the opposite angles of a cyclic quadrilateral is supplementary.