Additional Example 1
In the figure (a) below, A, B, C and D are points on the circle.
Compute the angles of the quadrilateral ABCD, and the angles between it's diagonals
Solution:
1. Consider ΔBPC. We get ∠BPC = [180-(30+50)] =[180-80] = 100o. (∵ sum of interior angles of a triangle is 180o) This is marked in fig(b)
2. ∠BPC and ∠BPA form a linear pair. So ∠BPA = 180 - ∠BPC = 180 -100 = 80o.
In the figure (a) below, A, B, C and D are points on the circle.
Solution:
1. Consider ΔBPC. We get ∠BPC = [180-(30+50)] =[180-80] = 100o. (∵ sum of interior angles of a triangle is 180o) This is marked in fig(b)
2. ∠BPC and ∠BPA form a linear pair. So ∠BPA = 180 - ∠BPC = 180 -100 = 80o.
3. ∠BPC and ∠APD are opposite angles, and are hence equal. So we get ∠APD = ∠BPC = 100o.
4 Similarly, ∠BPA and ∠DPC are opposite angles, and are hence equal. So we get ∠DPC = ∠BPA = 80o.
5. Consider arc BC in fig(c). It subtends ∠BAC (= 35o) on the alternate arc.
• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 35o.
6. Consider ΔPCD. We get ∠PCD = [180-(80+35)] =[180-115] = 65o. (∵ sum of interior angles of a triangle is 180o)
7. Consider arc CD in fig(c). It subtends ∠CBD (= 30o) on the alternate arc.
• The same arc subtends ∠CAD on the alternate arc. So ∠CAD = ∠CBD = 30o.
8. Consider ΔPAD. We get ∠ADP = [180-(100+30)] =[180-130] = 50o. (∵ sum of interior angles of a triangle is 180o)
■ Thus we get all the angles of the quadrilateral ABCD. Since it is a cyclic quadrilateral, we can do a check:
(i) Sum of opposite angles ∠BAD and ∠BCD = 35 + 30 + 50 + 65 = 180o.
(ii) Sum of opposite angles ∠ABC and ∠ADC = 65 + 30 + 50 + 35 = 180o.
Additional Example 2
In the figure (a) below, AB is a diameter of the circle.
CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Calculate ∠AEB
Solution:
1.Draw OC and OD as shown in fig(b). Consider the ΔOCD
• OC = OD (∵ radii of the same circle)
• Given that chord CD is equal to the radius.
• So ΔOCD is an equilateral triangle. All it's interior angles are equal to 60o.
2. Draw AD as shown in fig(c)
3. Consider the arc CD. It has a central angle ∠COD = 60o.
4. This same arc CD subtends ∠CAD on the opposite arc.
• So ∠CAD = 1⁄2×60 = 30o.
5. Now consider ∠ADB. AB is a diameter and D is a point on the semi-circle. So ADB = 90o.
6. Consider ΔAED. The ∠ADB that we considered above is an exterior angle of ΔAED
• Exterior angle of a triangle = sum of remote interior angles
• So ∠ADB = ∠EAD + ∠AED ⇒ 90 = 30 + ∠AED ⇒ ∠AED = 90 - 30 = 60o.
7. But ∠AED and ∠AEB are the same angles
• Thus the required ∠AEB = 60o
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