in triangle abc ad is the internal bisector of angle a if bd?
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In ΔABC, AD is the internal bisector of ∠A, meeting the side BC at D. If BD = 5 cm, BC = 7.5 cm, then AB : AC is.
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ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q. Given AB = 8 cm, AD = 5 cm, AC = 10 cm. When triangles are similar, their corresponding angles are equal.
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The opposite sides of a parallelogram are equal. ABCD is a parallelogram. To prove that AB = CD and AD = BC. ... Hence AB = CD and BC = AD (matching sides of congruent triangles).
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Theorem. and conversely, if a point D on the side BC of triangle ABC divides BC in the same ratio as the sides AB and AC, then AD is the angle bisector of angle ∠ A. This reduces to the previous version if AD is the bisector of ∠ BAC.
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In the given figure, ABCD is a cyclic quadrilateral, in which AC and BD are its diagonals. If ∠DBC = 60° and ∠BAC = 30°, then find ∠BCD. In the given figure, ABCD is a cyclic quadrilateral, in which AC and BD are its diagonals.
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The opposite sides of a parallelogram are equal. ABCD is a parallelogram. To prove that AB = CD and AD = BC.
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ABCD is a parallelogram X and Y are the midpoints of BC and CD respectively. Prove that ar(AXY) = 3/8ar(parallelogram ABCD).
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In the adjoining figure ABCD is a rhombus and EDC is an equilateral triangle. If ∠DAB = 48°, find. (1) ∠BEC (2) ∠DEB (3) ∠BFC. [Hint- ∠BCE = 48° + 60° = 180° , BC = EC]
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Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°. For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).
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ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively. Prove that PQRS is a parallelogram. ABCD is a quadrilateral in which P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively. ... Hence, PQRS is a parallelogram.
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ABCD is a parallelogram whose diagonals intersect each other at O with 90°. so it's a rhombus as diagonals of rhombus interest each other at right angles.
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The point at which the perpendicular bisectors of a triangle meet, or the circumcenter, is equidistant from the vertices of the triangle. On the other hand, angle bisectors simply split one angle into two congruent angles. Points on angle bisectors are equidistant from the sides of the given angle.
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The angles on the inside are called Interior angles. The sum of the interior angles of a triangle is always 180 degrees. The exterior angle is the angle between any side of a shape, and a line extended from the next side. The sum of an exterior angle and its adjacent interior angle is also 180 degrees.
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Intersection of diagonals (2) ABCD is a cyclic quadrilateral if and only if points P and Q are the midpoints of sides AB and CD. The intersection P may be internal or external to the circle. In the former case, the cyclic quadrilateral is ABCD, and in the latter case, the cyclic quadrilateral is ABDC.
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ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersect AB at P and DC at Q. Prove that: ar( POA) = ar( QOC)
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AD is perpendicular to the internal bisector of ∠ABC of △ABC. DE is drawn through D and parallel to BC to meet AC at E. If the length of AC is 12 cm, then the length of AB (in cm.) is. E will be mid point of AC.
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In figure ABCD is a rhombus whose diagonal AC makes an angle alpha with AB. If cos alpha = 2/3 & OB = 3cm, find the lengths of diagonals AC & BD.