Advertisements
Advertisements
प्रश्न
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angles of ΔDEF are 90° − `1/2 ∠A, 90° − 1/2 ∠B, 90° − 1/2` ∠C.
Advertisements
उत्तर

It is given that BE is the bisector of ∠B.
∴ ∠ABE = ∠B/2
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
⇒ ∠ADE = ∠B/2
Similarly, ∠ACF = ∠ADF = ∠C/2 (Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
`= (angleB)/2 + (angleC)/2`
`= 1/2(angleB + angleC)`
`= 1/2(180^@ - angleA)`
`= 90^@ - 1/2angleA`
Similarly, it can be proved that
`angleE = 90^@ - 1/2angleB`
`angleF = 90^@ - 1/2angleC`
APPEARS IN
संबंधित प्रश्न
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

In the figure, `square`ABCD is a cyclic quadrilateral. Seg AB is a diameter. If ∠ ADC = 120˚, complete the following activity to find measure of ∠ BAC.
`square` ABCD is a cyclic quadrilateral.
∴ ∠ ADC + ∠ ABC = 180°
∴ 120˚ + ∠ ABC = 180°
∴ ∠ ABC = ______
But ∠ ACB = ______ .......(angle in semicircle)
In Δ ABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°
∴ ∠BAC + ______ = 180°
∴ ∠ BAC = ______
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.

In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.

ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
