prove that the quadrilateral formed by the internal bisectors of any quadrilateral is cyclic
Answers
Answer:
Step-by-step explanation:
Let us consider ABCDABCD is a quadrilateral AH, BF, CF, DHAH,BF,CF,DH are bisector of \angle A, \angle B, \angle C , \angle D∠A,∠B,∠C,∠D we have to prove EFGHEFGH is cyclic quadrilateral to prove EFGHEFGH is cyclic we have to prove sum of one pair of opposite angle is 180^o180
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In \triangle le△le AEB \rightarrow \angle ABE+ \angle BAE+ \angle AEB=180^{o}AEB→∠ABE+∠BAE+∠AEB=180
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\angle AEB=180^{o}- \angle ABE- \angle BAE∠AEB=180
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−∠ABE−∠BAE
\angle AEB= 180^{o} (1/2 \angle B+ 1/2 \angle A)=180^{o}-1/2 ( \angle B+ \angle A)---(1)∠AEB=180
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(1/2∠B+1/2∠A)=180
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−1/2(∠B+∠A)−−−(1)
Now lines AHAH and BFBF intersect
So \angle FEH= \angle AEB∠FEH=∠AEB
\angle FEH=180^{o}-1/2 (\angle B+ \angle A)---(2)∠FEH=180
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−1/2(∠B+∠A)−−−(2)
Similarly \angle FGH = 180^{o}- \dfrac{1}{2} (\angle C+ \angle D)---(3)∠FGH=180
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−
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(∠C+∠D)−−−(3)
Add (2)(2) & (3)(3) \angle FEH+ \angle FGH =180^{o}- 1/2 (\angle B+ \angle C)+180^{o}-1/2 (\angle A+ \angle D)∠FEH+∠FGH=180
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−1/2(∠B+∠C)+180
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−1/2(∠A+∠D)
\angle FEH+ \angle FGH = 180^{o}+180^{o}-1/2 (\angle A+ \angle B+ \angle C + \angle D)∠FEH+∠FGH=180
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+180
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−1/2(∠A+∠B+∠C+∠D)
Since ABCDABCD is quadrilateral
Sum of angles =360=\angle A + \angle B+ \angle C +\angle D=360=∠A+∠B+∠C+∠D
\angle FEH+ \angle FGH=360- \dfrac{1}{2} (360)=360-180=180^{o}∠FEH+∠FGH=360−
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(360)=360−180=180
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\angle FEH+ \angle FGH=180^{o}∠FEH+∠FGH=180
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Thus EFGHEFGH is cyclic quadrilateral.
Since sum of one pair of opposite angle is 180^{o}180
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