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Three angles of a quadrilateral are equal. It is a parallelogram ?
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It need not be parallelogram we may have and
Diagonals of a parallelogram are perpendicular to each other. Is this statemetn true ? Give reason for your answer.
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No, diagonals of a parallelogram are not perpendicular to each other, they only bisect each other.
Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, find the lengths of AC and BD.
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As diagonals bisect each other in a parallelogram
The sides BA and DC of quadrilateral ABCD are produced as shown in the figure.
Prove that \(x + y = a + b \)
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(Linear pair)
In the figure. \(ABCD \) is a square. A line segment \(DX \) cuts the side \(BC \) at \(X \) and the diagonal \(AC \) at \(O \) such that \(\angle COD=105^o \). Find the of \(x \).
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In \(\triangle \)ABC,D and E are mid points of AB and AC. If AD = 3.5 cm ; AE = 4 cm ; DE = 2.5 cm, find the perimeter of \(\triangle \)ABC
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( and are mid points)
ABCD is a parallelogram, L and M are points on AB and DC respectively such that AL = MC. Prove that LM and BD bisect each other.
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In and
(VOA)
(alt. )
(AAS)
(cpct)
bisect and
l, m and n are three parallel lines intersected by transversal p and q such that l, m and n cut equal intercepts AB and BC on p. Show that l, m, n cut off equal intercept DE and EF on q also.
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is divided into two triangles is the mid point of
In is the mid point of
By mid point theorem, is the mid point of
In is the mid point of
is the mid point of .
(by mid point theorem)
In the figure. ABCD is a parallelogram BA is produced to E such that AE = AD. ED is produced to meet BC produced at F. Show that CD = CF.
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Corresponding angle
(given)
So,
In the given figure. PQRS is a parallelogram and line segments PA and RB bisect the angles P and R respectively. Show that PARB.
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(opposite angles of a parallelogram)
opposite angles of a quadrilateral are equal.
So, is a parallelogram
Hence
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that :
(i) \(\triangle APB\cong\triangle CQD \)
(ii) \(AP=CQ \)
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In and
(alt. )
(opposite sides)
(ASA)
(cpct)
ABCD is a quadrilateral in which P, Q, R and S are mid points of AB, BC, CD and DA respectively. Show that PQRS is a parallelogram.
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In
as and are the mid point …(1)
In
…(2)
So,
So, is a parallelogram
In the figure. it is given that BDEF and FDCE are parallelograms. Show that BD = CD.
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opposite sides of a parallelogram all equal
So, (Proved)
In a cycle quadrilateral \(PQRS \), if \(\angle P\angle R=50^o \). then find the measure of \(\angle P \) and \(\angle Q \)
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In \(\triangle \)ABC , AD is the median and DE  AB. Prove that BE is another median.
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Given is a median and
In
is mid point of and
By converse of mid point theorem, bisects
So is the median
In \(\triangle \)ABC, AB = 5 cm , BC = 8 cm and AC = 7 cm. If D and E are respectively mid–points of AB and BC, determine the length of DE. Give reasons.
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(By mid point theorem)
In the figure, it is given the BDEF and FDCE are parallelograms. If BD = 4 cm, determine CD.
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In parallelogram
….(1)
In parallelogram
….(2)
From equation (1) and (2)
So
In the figure, PQRS is a parallelogram. Find the values of x and y.
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(alternate angles)
(alternate angles)
In the figure, \(\angle AOB=90^o,AC=BC,OA=12\ cm \). and \(OC=6.5\ cm \) Find the measure of \(OB \).
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(given)
ABCD is a rhombus in which altitude from D on side AB bisects AB. Find the angles of the rhombus.
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In and
(common)
(each )
(given)
(SAS)
(cpct)
But (sides of rhombus are equal)
So,
is an equilateral
(linear pair)
(opposite angles of rhombus are equal)
In a triangle ABC, median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
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is a median
So, and also (given)
the diagonals of quadrilateral bisect each other
is a parallelogram
Show that the quadrilateral formed by joining the mid–points of sides of a square is also a square.
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…(1)
….(2)
…(3)
….(4)
From equation (1), (2), (3) and (4)
So FONM is a square
In the figure, P is the mid–point of side BC of a parallelogram ABCD such that \(\angle \)BAP = \( \angle\)DAP. Prove that AD = 2CD.
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…(1)
is a parallelogram
(Linear pair)
…(2)
In
(angle sum property)
From (1) and (2)
(opposite sides of a parallelogram) (Proved)
Two parallel lines l and m are intersected by transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
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(alt )
(as alt are equal)
Similarly
is a parallelogram
(Linear pair)
is a rectangle
In a parallelogram ABCD, bisector of \(\angle \)A. also, bisects BC at X. Prove that AD = 2 AB.
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(given)
( is the mid point)
(alt. )
(as )
(side opposite to equal angles are equal)
(opposite sides of parallelogram)
(Proved)
AD is the median of \(\triangle \)ABC. E is the midpoint of AD. BE produced meets AC at F. Show that \(\mathrm{AF=\dfrac{1}{3}AC} \).
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is parallel to and is the mid point of
By converse of mid point theorem, is the mid point of and …(1)
Since is the mid point and parallel to , by the converse of mid point theorem, is the mid point of and …(2)
From (1) and (2)
(Proved)
In a quadrilateral ABCD. AO and BO are the bisector of \(\angle \)A and \(\angle \)B respectively. Prove that \(\mathrm{\angle AOB=\dfrac{1}{2}(\angle C+\angle D)} \)
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In
…(1)
…(2)
From equation (1) and (2)
So,
In the figure. \(ABCD \) is a square. If \(\angle PQR=90^o \) and \(PB=QC=DR \), prove that \( QB = RC. PQ = QR,\angle QPR=45^o\).
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(sides of square)
…(2)
….(3)
But (given)
So,
In and
(given)
(each )
(Proved above)
(SAS)
(cpct)
In the figure, points A and B are on the same side of a line m. AD \(\bot \) m and meet m at D and E respectively. If C is the mid point of AB. prove that CD = CE.
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(by construction)
(given)
In and
(common)
(SAS)
(cpct)
ABCD is parallelogram. On diagonal BD are points P and Q such that DP = BQ. Show that APCQ is a parallelogram
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In and
(given)
(opposite sides)
(alt. )
(SAS)
(cpct)
Similarly we can prove that
So is a parallelogram
In the figure, diagonal BD of parallelogram ABCD bisects \(\angle \)B. Show that it bisects \(\angle \)D also.
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In and
(given)
(opposite sides)
(common)
(SAS)
(cpct)
Prove that quadrilateral formed by bisectors of the angles of a parallelogram is a rectangle.
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Since is a parallelogram
(Linear pair)
(V.O.A)
and
is a quadrilatral whose each angle is
Hence is a rectangle
In the figure, \(PS \) and \(RT \) are medians of \(\triangle PQR \) and \( SMRT\). Prove that \(QM=\dfrac{1}{4}PQ \).
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is the median of
In
is the mid point of
So, is the mid point of (by converse of mid point theorem)
In the figure, PQRS is a square. M is the midpoint of PQ and AB \(\bot \) RM . Prove that RA = RB.
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In and
(V.O.A)
( is the mid point)
(ASA)
(cpct)
In and
(each )
(common)
(SAS)
(cpct)
In the figure \(ABC \) is an isosceles triangle in which \(AB = AC. CD  AB \) and \( AD\) is bisectors of exterior \(\angle CAE \) of \(\triangle ABC \). Prove that \( \angle CAD=\angle BCA\) and \(ABCD \) is a parallelogram.
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In
…(1)
(ext. angle is equal to the sum of interior opposite angle)
But they form alternate
Also
is a parallelogram
P, Q, R, and S are respectively the mid–points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AD = BC. Prove that PQRS is a rhombus.
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…(1)
(By mid point theorem)
….(2)
Similarly in
In
All sides are equal so, is a rhombus
In a parallelogram ABCD, E and F are the mid– points of sides AB and CD respectively. Show that the line segment AF and EC trisects the diagonal BD.
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is a parallelogram
and
In
and
is a parallelogram
and
In and is the mid point of
So, is the mid point of
…(1)
In
is the mid point of
is the mid point of
….(2)
From (1) and (2)
This is the line segment trisects the diagonal
ABC is a triangle right angled at C. A line through the mid–point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) \(\mathrm{MD\bot AC} \)
(ii) \(\text D \) is mid–point of \(\text{AC} \)
(iii) \(\mathrm{MC=MA=\dfrac{1}{2}AB} \)
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(i) In
is the mid point of
is the mid point of (Line drawn through mid point of one sides of a triangle, parallel to another side, bisect the third side)
(ii) As and is transversal
(Interior angles on the same side of transversal are supplementary)
(Proved)
(iii) In and
( is the mid point of )
(each )
(common)
(SAS)
(cpct) ….(1)
In the figure, ABCD is a parallelogram. E is the mid–point of BC. DE and AB when produced meet at F. Prove that AF = 2AB.
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In and
(VOA)
(given)
(alternate angles)
(ASA)
(cpct)…(1)
(opposite sides of a parallelogram) …(2)
From (1) and (2)
ABCD is a a trapezium in which AB  CD and AD = BC Show that
(i) \(\angle A=\angle B \)
(ii) \(\angle C=\angle D \)
(iii) \(\triangle ABC\equiv\ \triangle BAD \)
(iv) Diagonal \(AC= \) diagonal \(BD \)
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(given)
(by construction)
is a parallelogram
(opposite sides)
But (given)
(angle opposite to equal sides are equal)
(Linear pair)
(construction)
(Sum of interior angles)
(ii)
But
So,
In and
(common)
(given)
(Proved above)
(SAS)
(cpct)
\(ABCD \) is a parallelogram in which \(\angle A=60^o \). If bisectors of \(\angle A \) and \(\angle B \) meet at \(P \), prove that \(AD=DP,PC=BC,DC=2AD \).
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(i) is a parallelogram
( bisects )
(alternate )
In
(ii) (sum of adjacent interior angles)
( is bisector of )
(alternate angles)
(iii)
Prove that in a triangle, the line segment joining the mid–points of any two sides is parallel to third side and is half of it.
Using the above, if P, Q, R are the mid–points of sides BC, AC and AB of \(\triangle \)ABC respectively and if PQ = 2.5 cm, QR = 3 cm, PR=3.5 cm, find the lengths of AB, BC and CA.
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Const: Draw to meet produced at
To prove:
Proof: (alternate interior)… (1)
( is mid point of ) … (2)
(vertically opposite) …(3)
(ASA congruence rule)
(cpct) or ….(4)
(cpct) ….(5)
But ( is mid point of )
Also (construction)
In quad
&
Therefore quadrilateral is a
or
(From equation (4))