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

Question

Bisectors of angles A, B and C of a triangle ABC intersect its
circumcircle at D, E and F respectively Prove that
angles of triangle DEF are 90o - A / 2,
90o - B / 2 and 90o - C / 2 Pls
explai it to me as fast as posssible also explai me the meaning of
circumcircle and circumcenter
thnxxs

Crazzy · Accepted Answer

Given: A Δ ABC in which bisectors of angles meet the circle at X, Y and Z respectively

To prove: $n n ∠ n n XYZ n n n = n 90 n ° n - n \frac{n}{n 1 n} n n n ∠ n B n n n n ∠ n n ZXY n n n = n 90 n ° n - n \frac{n}{n 1 n} n n n ∠ n A n n n n n n ∠ n n XZY n n n = n 90 n ° n - n \frac{n}{n 1 n} n n n ∠ n C n n n n$ Proof: We know that angles in a same segment are equal ∴ ∠BYX = ∠BAX ⇒ ∠BYX = $n n \frac{n}{n ∠ n A n} n n$ Also, ∠BYZ = ∠BCZ ⇒ ∠BYZ = $n n \frac{n}{n ∠ n C n} n n$ ∴ ∠BYX + ∠BYZ = $n n \frac{n}{n ∠ n A n} n + n \frac{n}{n ∠ n C n} n n$ ⇒∠XYZ = $n n \frac{n}{n 1 n} n n (n ∠ n A n n n + n ∠ n C n n n) n n n$ ⇒∠XYZ = $n n \frac{n}{n 1 n} n n (n 180 n ° n n n - n ∠ n B n n n) n n n n (n âˆµ n ∠ n A n n n + n ∠ n B n n n + n ∠ n C n n n = n 180 n ° n n n n n n n) n n n$ ⇒∠XYZ = 90° – $n n \frac{n}{n 1 n} n n ∠ n B n n n$ Now, AZ subtends angles ∠ ACZ and ∠ AXZ at points C and X in the same segment ∴ ∠ACZ = ∠AXZ ⇒ ∠AXZ = $n n \frac{n}{n 1 n} n n ∠ n C n n n$ Also,∠AXY = ∠ABY ⇒ ∠AXY = $n n \frac{n}{n 1 n} n n ∠ n B n n n$ ∴ ∠AXZ + ∠AXY = $n n \frac{n}{n ∠ n C n} n + n \frac{n}{n ∠ n B n} n n$ ⇒ ∠ZXY = $n n \frac{n}{n 1 n} n n (n ∠ n B n n n + n ∠ n C n n n) n n n$ ⇒∠ZXY = $n n \frac{n}{n 1 n} n n (n 180 n ° n n n - n ∠ n A n n n) n n n$ ⇒∠ZXY =90° – $n n \frac{n}{n 1 n} n ∠ n A n n$ Similarly, we can prove that∠ XZY = 90° – $n n \frac{n}{n 1 n} n n n n ∠ n C n n n n n$ Here's the link as well ( this was answer given by the expert ) :) :)