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Q) 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°– \frac{A}{2}, 90°– \frac{B}{2} and 90°– \frac{C}{2}.

9th Class Maths – NCERT Important Questions

Ans:

Step 1: Let’s make a diagram for better understanding of the question:

Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively

Here, ABC is a triangle inside the circle. AD, BE and DF are bisectors of ∠ A, ∠ B an ∠ C respectively.

Step 2: Let’s connect AE. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively

We know that angles in same segment are always equal,

∴ ∠ ABE = ∠ ADE

∵ BE is bisecor of ∠ B

∴ ∠ ABE = ∠ ADE = \frac{B}{2}   ………. (i)

Step 3: Let’s connect AF. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively

∵ angles in same segment are always equal,

∴ ∠ ACF = ∠ ADF

∵ CF is bisecor of ∠ C

∴ ∠ ACF = ∠ ADF = \frac{C}{2} ………. (ii)

Step 4: Now, value of ∠ D = ∠ ADE + ∠ ADFBisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively

∴ ∠ D = \frac{B}{2} + \frac{C}{2}

∴ ∠ D = \frac{1}{2} (B + C) ….. (iii)

Step 5: In Δ ABC, ∠ A + ∠ B + ∠ C = 1800

∴ ∠ B + ∠ C = 1800 – ∠ A ………. (iv)

Now, by substituting equation (iv) in equation (iii), we get:

∴ ∠ D = \frac{1}{2} (B + C) .

∴ ∠ D = \frac{1}{2} (1800 – ∠ A)

∠ D = 900\frac{A}{2}

Similarly, ∠ E = 900\frac{B}{2}

and ∠ F = 900\frac{C}{2}

Hence Proved!

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