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If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent. Is the statement true ? Why?
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The statement is true because the triangles will be congruent either by ASA rule or AAS rule. This is because two angles and one side are sufficient to construct two congruent triangles.
In the two triangles ABC and DEF, AB = DE, and AC = EF. Name two angles from the triangles that must be equal so that the two triangles are congruent. Give reason for your answer.
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Given
The two angle are
(SAS)
(ASS)
(SSA)
Prove that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle.
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In ,
is the longest side
In ,
is the longest side
In
is the longest side
So adding all the three we get that perimeter of a triangle is greater than sum of its three altitudes
AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2 AD. Give reason for your answer.
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is a median,
(Proved)
M is a point on side BC of a triangle ABC such that AM is the bisector of \(\angle \)BAC . Is it true to say that AB + BC + CA > 2 AM ? Give reason for your answer.
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In
(Sum of two sides is greater than third side)
Adding equation (1) and (2)
Arrange the sides of \(\triangle \)ABC in ascending order of lengths.
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\(\)\angle A+\angle B=110^o\ โฆ(1)\\ \\ \angle B+2C=130^o\ โฆ.(2)\\ \\ \angle C\angle A=20^o\\ \\ \angle C=\angle A+20^o\\ \\ \angle C=180110=70^o\\ \\ \angle A=7020=50\\ \\ \angle B=180(70+50)=60^o\\ \\ \angle A<\angle B<\angle C\\ \\ BC<AC
Is it possible to construct a triangle with lengths of its sides as 8 cm, 9 cm, and 2 cm,? Give reason for your answer.
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Yes it is possible as sum of two sides is greater than the third side
In an equilateral triangle \(ABC\), if \(AD \) is a median, then prove that \(\angle ADC=90^o \).
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In and
(given)
But (linear pair)
and
In the figure, PR = QR, \(\angle \)PRA = \(\angle \)QRB and \(\angle \)BPR = \(\angle \)AQR.Prove that BP = QA.
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In and
(given)
(given)
(addition of same angles with equal angle will also be equal)
(AAS)
(cpct)
In \(\triangle \)ABC, AD is perpendicular bisector of BC. Show that \(\triangle \)ABC is an isosceles triangle in which AB = AC.
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In and
(common)
(each )
(given)
(SAS)
(cpct)
Therefore ABC is isosceles triangle in which
Prove that in an isosceles triangle, angles opposite to equal sides are equal.
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In and
(given)
(each )
(common)
(RHS)
(cpct)
In the figure. \(\angle ABD=\angle ACE \) and \(AB=AC \). Prove that \(\triangle ABD\cong\triangle ACE \)
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In and
(given)
(common)
(given)
(AAS)
In \(\triangle \)ABC, AB = AC. D is a point inside \(\triangle \)ABD such that BD = DC. Prove that \(\angle \)ABD = \(\angle \)ACD.
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In and
(given)
(common)
(SSS)
(cpct)
In the figure X and Y are two points on equal sides AB and AC of a \(\triangle \)ABC such that AX = AY. Prove that XC = YB.
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In and
(given)
(given)
(common)
(SAS)
(cpct)
In the figure, ABCD is a square and P is the midpoint of AD, BP and CP are joined. Prove that \(\angle \)PCB = \(\angle \)PBC
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In and
(given)
(each )
( is a square)
(cpct)
(Proved)
In the figure, the diagonal AC of quadrilateral ABCD bisects \(\angle \)BAD and \(\angle \)BCD . Prove that BC = CD.
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In and
(given)
(given)
(common)
(AAS)
(cpct)
In the figure, AB > AC, BO and CO are the bisectors of \(\angle \)B and \(\angle \)C respectively. Show that OB > OC.
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and are bisectors
is oppsite to and is opposite to
So (Proved)
In the figure, PQ = PR and \(\angle \)Q = \(\angle \)R, Prove that PQS \(\cong\ \triangle \)PRT
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In and
(given)
(given)
(common)
(AAS)
(Proved)
In the figure. AB \(\bot \) AC, DE \(\bot \) DF such that BA = DE and BF = EC. Show that \( \triangle\)ABC \(\cong\ \triangle \)DEF
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In and
(given)
(each )
(RHS)
In the figure, D is any point on the side BC of a triangle ABC. Prove that AB + BC + CA > 2AD.
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In
(Sum of two sides is greater than third side) โฆ(1)
In
โฆ(2)
Adding equation (1) and (2)
(Proved)
In the figure, \(AB = AC\) and \(\angle1=\angle2 \). Prove that \(\angle PBC=\angle PCB \).
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In and
(given)
(given)
(common)
(SAS)
(cpct)
(Angle opposite to equal sides are equal)
In the figure, two lines AB and CD interect each other at O such that BCDA and BC=DA. Show that O is the midpoint of both the line segments AB and CD.
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In and
(given)
(alternate angle)
(alternate angle)
(ASA)
(cpct)
(cpct)
In the figure, AD is the bisector of \(\angle \)BAC . Prove that AB > BD.
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Since is the bisector of
(exterior angle is greater than each of the opposite interior angle)
In the figure, l  m and M is the midpoint of a line segment AB. Show that M is also the midpoint of any line segment CD having its end points on l and m respectively.
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In and
(given)
(V.O.A)
(alt. )
(AAS)
(cpct)
In the figure, D is any point on the base BC produced of an isosceles triangle \(\triangle \)ABC . Prove that AD > AB.
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( ABC is an isosceles )
In
ext.
In right triangle ABC, \(\angle C=90^o \), M is midpoint of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that
(i) \(\triangle AMC\cong\triangle BMD \)
(ii) \(\triangle DBC\cong\triangle ACB \)
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In and
(given)
(VOA)
(alt )
(AAS)
(ii) In and
(common)
(common)
(common)
(AAS)
In the figure. PR > PQ and PS bisects \(\angle \)PQR . Prove that \(\angle \)PSR > \( \angle\)PSQ .
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In
In
Adding to both sides
In the figure, the perpendicular AD, BE and CF drawn from the vertices A,B and C respectively of \(\triangle \)ABC are equal. Prove that the triangle is an equilateral triangle.
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In and
(given)
(common)
(ASA)
In and
(Common)
(ASA)
(cpct)
(Proved)
Prove that medians of an equilateral triangle are equal.
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are the medians of
and
( each)
(SAS)
Similarly
So
In the figure, \(\triangle \)LMN is an isosceles triangle with LM = LN and LP bisects \(\angle \)NLQ. Prove that LP  MN.
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In
From (1) and (2)
But they form corresponding angle
So,
In the figure. D is a point on side BC of \(\triangle \)ABC such that AD = AC. Show that AB > AD.
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(exterior angle property)
(Proved)
O is any point in the interior of \(\triangle \)ABC. Show that OB + OC < AB + AC
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In
In
โฆ(1)
In
โฆ(2)
In
….(3)
….(4)
In
…(5)
Adding (4) and (5)
In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.
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In and
(by construction)
(Common)
(SAS)
(cpct),
ABC is a right angled triangle with AB = AC. Bisector of \(\angle \)A meets BC at D. Prove that BC = 2AD.
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So โฆ.(1)
Similarly
….(2)
Adding equation (1) and (2)
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC).
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In โฆ(1)
In โฆ.(2)
In โฆ.(3)
In โฆ.(4)
In the figure, A is a point equidistant from two lines \(l_1 \) and \(l_2 \) intersecting at a point P, show that AP bisects the angle between \(l_1 \) and \(l_2 \).
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In and
(given)
(Common)
(each )
(RHS)
(cpct)
(Proved)
In a triangle PQR, PR > PQ and PS is the bisector of \(\angle \)QPR. Prove that \(\angle \)PSR > \(\angle \)PSQ.
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Let
In
โฆ(1)
In ,
….(2)
In the figure, \(AC = AE, AB = AD \) and \(\angle BAD=\angle EAC \). Show that \(BC=DE \).
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In and
(given)
(given)
(SAS)
(cpct)
In the figure, show that 2(AC + BD) > (AB + BC + CD + DA)
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In
โฆ.(1)
(Sum of two sides is greater than the third side)
In
โฆ(2)
In ,
….(3)
In
….(4)
\(O \) is a point in the interior of \(\triangle PQR \). Prove that \(OP + OQ + OR >\dfrac{1}{2}(PQ + QR + PR) \)
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In ,
โฆ(1)
In
โฆ.(2)
In
โฆ.(3)
Adding 3 equation we get
Show that perimeter of a triangle is greater than the sum of its medians.
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We know that sum of any two sides of a triangle is greater than twice the median
Adding we get
AB and CD are respectively the smallest and the longest sides of a quadrilateral ABCD as shown in the figure. Prove that \(\angle A>\angle C \) and \(\angle B>\angle D \)
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In ,
โฆ(1)
In ,
is the largest side
โฆ.(2)
Adding (1) and (2)
Similarly by joining we can prove that
In the figure, if two isosceles triangles have a common base, prove that the line segment joining their vertices bisects the common base at right angles.
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In and
(sides of one isosceles )
(sides of one isosceles )
(common)
(cpct)
(cpct)
But (Linear pair)
In the figure, \(AC=BC,\ \angle DCA=\angle ECB \) and \(\angle DCB=\angle EAC \). Prove that (i) \(\triangle DBC\cong\triangle EAC \) ; (ii) \(DC=EC \) and \(BD=AE \).
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In and
(given)
(given)
(AAS)
(cpct)
(cpct)
In the figure, BAPQ, CA RS and BP=RC. Prove that (i) BS = PQ ; (ii) RS = CQ.
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Similarly
In and
(ASA)
(cpct)
(cpct)
In a right triangle ABC, right angled at C, M is the mid point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that :
(i) \(\triangle AMC\cong\triangle BMD \)
(ii) \(\angle DBC=90^o \)
(iii) \(\triangle DBC\cong\triangle ACB \)
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In and
(given)
(VOA)
(SAS)
(cpct)
(cpct)
But they form linear pair
So,
In and
(Common)
(Proved above)
(Proved above)
(RHS)
In the figure, \(\angle \)QPR = \(\angle \)PQR and M and N are respectively points on sides QR and PR of \(\triangle \)PQR , such that QM = PN. Prove that OP = OQ, where O is the point of intersection of PM and QN.
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In and
(given)
(given)
(SAS)
(cptc)
In and
(given)
(VOA)
(Proved above)
(AAS)
(cpct)
In the figure, PQ and RS are perpendicular to QS,QA = BS and PB = AR. Prove that \(\angle \)QPB = \( \angle\)SRA
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In and
(given)
(given)
(each )
(RHS)
(cpct)
In the figure, if AD is the bisector of \(\angle \)BAC , then prove that :
(i) AB > BD
(ii) AC > CD
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( is bisector of )
(exterior angle is greater than each interior opposite angle
(Side opposite to greater angle is larger)
(Side opposite to greater angle is greater)