Question

Prove that isosceles triangle is always cyclic

Answer 1

To prove that any given quadrilateral is cyclic, we need to prove that its opposite angles are supplementary (i.e. they add up to 180˚).
 Usually, in ΔADF and ΔBCE,
∠AFD = ∠BEC (by construction - both are right angles at the feet of the perpendiculars)AD = BC (property of trapezium)AF = BE ( perpendicular distance between two parallel lines remains same irrespective of the point of measurement along the line )Thus, ΔADF ≅ ΔBCE by RHS ( Right angle - Hypotenuse - Side ) congruency.
 
Now,
∠ADF = ∠BCE ( By CPCT i.e. Corresponding Parts of Congruent Triangles )Since ∠ADF is same as ∠ADC and ∠BCE is same as ∠BCD,
∠ADC = ∠BCD (equation 1)
Also,
∠CBE = ∠DAF ( By CPCT )Adding the right angles ∠ABE and ∠BAF to the above angles,
∠CBE + ∠BAF = ∠CBE + ∠ABE
Thus, ∠ABC = ∠BAD (equation 2)

So, adding equations 1 and 2,
∠ADC + ∠ABC = ∠BCD + ∠BAD
Since the sum of all the angles in a quadrilateral is 360˚,
∠ADC + ∠ABC + ∠BCD + ∠BAD = 360˚
2 (∠ADC + ∠ABC) = 2 (∠BCD + ∠BAD) = 360˚
∠ADC + ∠ABC =∠BCD + ∠BAD = 180˚

Since the opposite angles are supplementary, an isosceles trapezium is a cyclic quadrilateral.
Hence proved.

Hope that helps :)

Answer 2

Answer:

Every triangle is cyclic.

Step-by-step explanation:

Triangle ABC. Line l1 is perpendicular bisector of AB, line l2 is perpendicular bisector of AC. Hence their intersection point is equidistant from A,B and C, so there exists a circle through point A, B and C. The intersection point is the center of the circle.