1. You are given a circle. Give a construction to find its center.
2. Prove that the quadrilateral formed by the internal angle bisectors of any
quadrilateral is cyclic.
3. If non-parallel sides of a trapezium are equal, prove that it is cyclic.
2. Prove that the quadrilateral formed by the internal angle bisectors of any
quadrilateral is cyclic.
3. If non-parallel sides of a trapezium are equal, prove that it is cyclic.
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2.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 + ½ ∠A + ½ ∠ B = 180°
∠AEB = 180° – ½ (∠A + ∠ B) -------- (2)
From (1) and (2),
∠HEF = 180° – ½ (∠A + ∠ B) --------- (3)
Similarly, ∠HGF = 180° – ½ (∠C + ∠ D) -------- (4)
From 3 and 4,
∠HEF + ∠HGF = 360° – ½ (∠A + ∠B + ∠C + ∠ D)
= 360° – ½ (360°)
= 360° – 180°
= 180°
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180°.
3. Let the trapezium ABCDABCD be isosceles and the parallel sides be ABAB and CDCD. The equal length non-parallel sides are BCBC, DADA. The angles ∠DAB∠DAB and ∠CDA∠CDA are supplementary in the parallel lines so ∠DAB+∠CDA=π∠DAB+∠CDA=π.
The trapezium is symmetric so the angles ∠DAB∠DAB and ∠ABC∠ABC are equal.
So we have ∠ABC+∠CDA=π∠ABC+∠CDA=π, i.e. we have opposite angles in the trapezium add up to ππ. This is necessary and sufficient for a quadrilateral to be cyclic.
For a visual illustration, draw a isosceles trapezium, then draw the perpendicular bisectors of the two non-parallel sides. Think about the distance from the point where the bisectors intersect and each of the vertices of the trapezium