PRESENTED BYPRESENTED BY

PRESENTED BYPRESENTED BY

Shivam Shivam GoelGoelIX-EIX-E

Examination Roll No.-159212Examination Roll No.-159212

QUADRILATERAL :QUADRILATERAL :A quadrilateral is a geometrical figure which has four sides, four angles, four vertices, and two diagonals. The sum of all angles of a quadrilateral is 360

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C D

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There are actually six types of quadrilaterals. They are as follows: Trapezium Parallelogram Rectangle Rhombus Square Kite

TRAPEZIUM :If in a quadrilateral one pair of opposite sides are equal then the given quadrilateral is called a TRAPEZIUM.TRAPEZIUM.E.g. : In the above figure if AB is parallel to CD then the figure is a quadrilateral.

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CD

PARALLELOGRAM:If in a quadrilateral both the pairs of opposite sides are parallel , then the given quadrilateral is a PARALLELOGRAM.PARALLELOGRAM.

E.G.- In the above figure if AB is parallel to CD and AD is parallel to BC then the figure is a parallelogram.

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C D

RECTANGLE:If in a quadrilateral one of its angles is a right angle then the quadrilateral is a rectangle.E.G. : If in the above figure angle A is a right angle then the figure is a rectangle.

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CD

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D CRHOMBUS:If in a parallelogram all sides are equal, then the parallelogram is a RHOMBUS.E.G. : If in the above figure AB=BC=CD=DA, then it is rhombus.

SQUARE:A parallelogram whose one angle is a right angle and all the sides are equal, then it is called a SQUARE.E.G.: If in the above figure AB=BC=CD=DA, and angle B is a right angle, then the given figure is a square.

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CD

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B

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D

KITE:KITE:In a quadrilateral if two pairs of adjacent sides are equal. Then it is not a parallelogram. It is called a KITE.E.G. : If in the above figure AB = AD and BC = CD, then it is not a parallelogram. It is a kite.

A square is a rectangle and also a rhombus. A parallelogram is a trapezium. A kite is not a parallelogram. A trapezium is not a parallelogram. A rectangle or a rhombus is not a square .

The sum of angles of a quadrilateral is 360 degrees. A diagonal of a parallelogram divides it into two congruent triangles. In a parallelogram opposite sides are equal. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. In a parallelogram opposite angles are equal.If in a quadrilateral each pair of opposite angles is equal, then it is a parallelogram.

The diagonals of a parallelogram bisect each other. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.The line segment joining the mid-points of two sides of a triangle is parallel to the third side. The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Theorem : Sum of angles of a quadrilateral is 360

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Given: A quadrilateral ABCD.To prove: angles A + B+ C+ D= 360.Construction: Join A to C.Proof: In triangle ABC, angle CAB + angle ACB + angle CBA = 180. (A.S.P) – 1 In triangle ACD, angle ADC + angle DCA + angle CAB = 180 (A.S.P) -2 Adding 1 and 2 angles CAB+ACB+CBA+ADC+DCA+CAD=180+180 angles (CAB+BAC)+ABC+(BCA+ACD)+ADC= 360. Therefore, angles A+B+C+D=360.

THEOREM: The diagonal of a parallelogram divides it into two congruent triangles.

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Given: A parallelogram ABCD and its diagonal AC.To prove: Triangle ABC is congruent to triangle ADCConstruction: Join A to C.Proof: In triangles ABC and ADC, AB is parallel to CD and AC is the transversal Angle BAC = Angle DCA (alternate angles) Angle BCA = Angle DAC (alternate angles) AC = AC (common side) Therefore, triangle ABC is congruent to triangle ADC by ASA rule. Hence Proved.

Given: A parallelogram ABCD and its diagonal AC.To prove: Triangle ABC is congruent to triangle ADCConstruction: Join A to C.Proof: In triangles ABC and ADC, AB is parallel to CD and AC is the transversal Angle BAC = Angle DCA (alternate angles) Angle BCA = Angle DAC (alternate angles) AC = AC (common side) Therefore, triangle ABC is congruent to triangle ADC by ASA rule. Hence Proved.

THEOREM: In a parallelogram , opposite sides are equal. D

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CGiven: A parallelogram ABCD.To Prove: AB = DC and AD = BCConstruction: Join A to CProof: In triangles ABC and ADC, AB is parallel to CD and AC is the transversal. Angle BAC = Angle DCA (alternate angles) Angle BCA = Angle DAC (alternate angles) AC = AC (common side) Therefore, triangle ABC is congruent to triangle ADC by ASA rule. Now AB = DC and AD = BC (C.P.C.T) Hence Proved.

THEOREM: If the opposite sides of a quadrilateral are equal, then it is a parallelogram.

A B

CDGiven: A quadrilateral ABCD in which AB=CD & AD=BCTo Prove: ABCD is a parallelogram.Construction: Join A to C.Proof: In triangle ABC and triangle ADC , AB = CD (given) AD = BC (given) AC = AC (common side)Therefore triangle ABC is congruent to triangle ADC by SSS ruleSince the triangles of a quadrilateral are equal, therefore it is a parallelogram.

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THEOREM: In a parallelogram opposite angles are equal.

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DGiven: A parallelogram ABCD.To prove: Angle A = Angle C & angle B=angle DProof: In the parallelogram ABCD, Since AB is parallel to CD & AD is transversal angles A+D=180 degrees (co-interior angles)-1 In the parallelogram ABCD, Since BC is parallel to AD & AB is transversal angles A+B=180 degrees (co-interior angles)-2 From 1 and 2, angles A+D=angles A+B. angle D= angle B.Similarly we can prove angle A= angle C.

THEOREM: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

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CD

A

Given: In a quadrilateral ABCD angle A=angle C & angle B=angle D.To prove: It is a parallelogram.Proof: By angle sum property of a quadrilateral, angles A+B+C+D=360 degrees angles A+B+A+B=360 degrees (since, angle A=C and angle B=D) 2angle A+ 2angle B=360 degrees 2(A+B)=360 degrees angles A+B= 180 degrees. (co-interior angles.) Therefore, AD is parallel to BCSimilarly’ we can prove AB is parallel to CD. This shows that ABCD is a parallelogram.

THEOREM: The diagonals of a parallelogram bisect each other.

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Given: A parallelogram ABCDTo prove: AO= OC & BO= OD.Proof: AD is parallel to BC & BD is transversal. angles CBD= ADB (alternate angles) AB is parallel to CD & AC is transversal. angles DAC= ACB (alternate angles) Now, in triangles BOC and AOD, CBD=ADB DAC=ACB BC=AD (opposite sides of a parallelogram)Therefore, triangle BOC is congruent to triangle AOD by ASA rule.Therefore, AO=OC & BO=OD [C.P.C.T]This implies that diagonals of a parallelogram bisect each other.

THEOREM: If the diagonals of a quadrilateral bisect each other then it is a parallelogram.

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CD

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Given: In a quadrilateral ABCD, AO = OC & BO = ODTo Prove: ABCD is a parallelogram.Proof: In triangles AOD & BOC AO = OC (given) BO = OD (given) angles AOD = BOC (vertically opposite angles)Therefore, triangle BOC is congruent to triangle AOD by SAS rule Therefore angle ADB = CBD & angle DAC = ACB (C.P.C.T)Since alternate angles are equal, AD is parallel to BC.Similarly, we can prove AB is parallel to CD.This proves that ABCD is a parallelogram .

THEOREM: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

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CD

A

Given: In a quadrilateral ABCD, AB is parallel to CD AB = CDTo prove: ABCD is a parallelogram.Construction: Join A to C.Proof: In triangles ABC & ADC, AB = CD ( given) angle BAC = angle DCA (alternate angles.) AC= AC ( common)Therefore, triangle ABC is congruent to triangle ADC by SAS rule.Therefore, angle ACB=DAC and AD=BC [C.P.C.T]Since, AD is parallel to BC and AD=BC,ABCD is a parallelogram.

THEOREM: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given: A triangle ABC in which D and E are the mid- points of AB and Ac respectively.To prove: DE is parallel to BC & DE=1/2BCProof: In triangles AED and CEF AE = CE (given) ED = EF (construction) angle AED = angle CEF (vertically opposite angles) Therefore, triangle AED is congruent to triangle CEF by SAS rule.Thus, AD=CF [ C.P.C.T] angle ADE = angle CFE [C.P.C.T]

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Now, AD= CFAlso, AD = BDTherefore, CF = BDAgain angle ADE = angle CFE (alternate angles)This implies that AD is parallel to FCSince, BD is parallel to CF (since, AD is parallel to CF and BD=AD).And, BD=CFTherefore, BCFD is a parallelogram.Hence, DF is parallel to BC and DF=BC (opposite sides of a parallelogram).Since, DF=BC; DE=1/2 BC Since, DE=DF (given) Therefore, DE is parallel to DF.

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Given: E is the mid- point of AB, line ‘l’ is passing through E and is parallel to BC and CM is parallel to BA.To prove: AF=CFProof: Since, Cm is parallel to BA and EFD is parallel to BC, therefore BEDC is a parallelogram.BE= CD( opposite sides of a parallelogram)But, BE = AE, therefore AE=CD.In triangles AEF & CDF: angle 1=2 (alt.angles) angle 3=4 (alt.angles) AE=CD (proved)Therefore,triangle AEF is congruent to CDF(ASA)AF=CF [C.P.C.T]. Hence, proved.

THEOREM: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.