Prove that the quadrilateral formed (if possible) by the internal angl...
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 + 1/2 ∠A + 1/2 ∠ B = 180degree1/2
∠AEB = 1/2 – 1/2 (∠A + ∠ B) -------- (2)
From (1) and (2),
∠HEF = 1/1/2 (∠A + ∠ B) --------- (3)
Similarly, ∠HGF = 1/2 (∠C + ∠ D) -------- (4)
From 3 and 4,
∠HEF + ∠HGF = 1/1/2 (∠A + ∠B + ∠C + ∠ D)
= 1/1/2 (1/2)
= 1/2 – 1/2
= 180degree1/2
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180 degree.]
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Prove that the quadrilateral formed (if possible) by the internal angl...
Introduction
The quadrilateral formed by the internal angle bisectors of a given quadrilateral is known to be cyclic. This means that the vertices of this quadrilateral lie on a single circle. Here’s a detailed proof of this property.
Angle Bisectors and Their Properties
- Let the quadrilateral be \(ABCD\).
- The internal angle bisectors of angles \(A\), \(B\), \(C\), and \(D\) intersect at four points: \(P\), \(Q\), \(R\), and \(S\).
Key Concept: Opposite Angles
- For a cyclic quadrilateral, the sum of opposite angles is \(180^\circ\).
- We will show that the sum of opposite angles for the quadrilateral \(PQRS\) equals \(180^\circ\).
Proof of Cyclic Nature
1. Angle Relations:
- The angles formed at point \(P\) from bisectors of angles \(A\) and \(B\) are \( \frac{A}{2} + \frac{B}{2} \).
- Similarly, at \(R\), the angles are \( \frac{C}{2} + \frac{D}{2} \).
2. Summing Opposite Angles:
- \( \angle P + \angle R = \left( \frac{A}{2} + \frac{B}{2} \right) + \left( \frac{C}{2} + \frac{D}{2} \right) \)
- This simplifies to \( \frac{A + B + C + D}{2} \).
3. Using Quadrilateral Properties:
- The sum of the angles in any quadrilateral \(ABCD\) is \(360^\circ\).
- Therefore, \(A + B + C + D = 360^\circ\).
4. Final Calculation:
- Thus, \( \angle P + \angle R = \frac{360^\circ}{2} = 180^\circ \).
- Similarly, \( \angle Q + \angle S = 180^\circ\).
Conclusion
Since the sum of opposite angles in quadrilateral \(PQRS\) equals \(180^\circ\), it confirms that the quadrilateral formed by the internal angle bisectors of any quadrilateral is indeed cyclic.
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