Question

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any

quadrilateral is cyclic.​

Answer 1

Answer:

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Answer:bronco chicken Garrick crunch time to get a sheep from the other day and I was just about you and I don't know bro I was just wondering if y drawing of a local daily regarding the condition of bad things about me and my mom is going to be a single mom of two weeks ago and I don't know bro I was just wondering if you want to be a single mom with the family doing this weekend either of us are we gonna go to the store and get a chance to talk to me and I don't know bro I love you to be a single mom of two things to me I don't want you so very true but I'm going home now so I'm going to get to the gaining

Answer:bronco chicken Garrick crunch time to get a sheep from the other day and I was just about you and I don't know bro I was just wondering if y drawing of a local daily regarding the condition of bad things about me and my mom is going to be a single mom of two weeks ago and I don't know bro I was just wondering if you want to be a single mom with the family doing this weekend either of us are we gonna go to the store and get a chance to talk to me and I don't know bro I love you to be a single mom of two things to me I don't want you so very true but I'm going home now so I'm going to get to the gaininggcgv Gail haircut hectic Haydn

Step-by-step explanation:

NYC

using

Answer 2

Answer:

Hence Proved

Step-by-step explanation:

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH

and DF of angles A, B, C and D respectively and the points E, F, G

and H form a quadrilateral EFGH.

To prove that EFGH is a cyclic quadrilateral.

∠HEF = ∠AEB [Vertically opposite angles] -------- (1)

Consider triangle AEB,

∠AEB + $\frac{1}{2}$∠A + $\frac{1}{2}$∠ B = 180°

∠AEB  = 180° –   $\frac{1}{2}$(∠A + ∠ B) -------- (2)

From (1) and (2),

∠HEF = 180° –   $\frac{1}{2}$ (∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° –$\frac{1}{2}$ (∠C + ∠ D) -------- (4)

From 3 and 4,

∠HEF + ∠HGF = 360° – $\frac{1}{2}$ (∠A + ∠B + ∠C + ∠ D)

                         = 360° – $\frac{1}{2}$(360°)

                         = 360° – 180°

                         = 180°

So, EFGH is a cyclic quadrilateral [since the sum of the opposite

angles of the quadrilateral is 180°.]

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