Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th

Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th
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An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

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ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

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Prove that a triangle must have atleast two acute angles. -Maths 9th

Description : Prove that a triangle must have atleast two acute angles. -Maths 9th

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Last Answer : We have ∠BED = ∠BAD (Angles in the same segment) ⇒ ∠BED = 1/2∠A ...(i) Also, ∠BEF = ∠BCF (Angles in the same segment) ⇒ ∠BEF = 1/2∠C ...(ii) From (i) and (ii) ∠BED + ∠BEF = 1/2∠A + 1/2∠C ∠DEF ... ∠A + ∠C) ⇒ ∠DEF = 1/2(180° - ∠B) (Since, ∠A + ∠B + ∠C = 180°) ⇒ ∠DEF = 90° - 1/2∠B

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If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

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If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

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Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

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If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

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Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

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Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

If two parallelogram PQAD and PQBC arw on the opposite sides of PQ prove that ABCD is a paralellogram -Maths 9th

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Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -Maths 9th

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Prove that in a triangle,other than an euilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle. -Maths 9th

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2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

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Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

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Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

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Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

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Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

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If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

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If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

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Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

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Prove that altitude of a triangle are equal -Maths 9th

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Prove that altitude of a triangle are equal -Maths 9th

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If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

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Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Description : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

Let A, B, C be the angles of a plain triangle (A/2) = (1/3 ), tan (B/2) = (2/3). Then tan (C/2) is equal to -Maths 9th

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The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC