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    quadrilateral quadrilateral Presentation Transcript

    • Quadrilaterals
    • Quadrilaterals Warm Up Problem of the Day Lesson Presentation
    • Warm Up Identify the triangle based on the given measurements. 1. 3 ft, 4 ft, 6 ft 2. 46°, 90°, 44° 3. 7 in., 7 in., 10 in. scalene right isosceles
    • Problem of the Day One angle of a parallelogram measures 90°. What are the measures of the other angles? all 90°
    • Learn to name and identify types of quadrilaterals.
    • Some quadrilaterals have properties that classify them as special quadrilaterals. Quadrilateral just means "four sides" (quad means four, lateral means side). Any four-sided shape is a Quadrilateral. But the sides have to be straight, and it has to be 2-dimensional.
    • 1.Parallelogram Properties Quadrilateral classification: Quadrilaterals are classified according to the number of pairs of each parallel sides.  If a quadrilateral does not have any pair of parallel sides, it is called a TRAPEZIUM. If a quadrilateral has only one pair of parallel lines, it is called TRAPEZOID.
    • If a quadrilateral has two pairs of parallel lines, it is called PARALLELOGRAM. A quadrilateral with opposite sides parallel and equal is a parallelogram . Properties:- • A diagonal of a parallelogram divides it into two congruent triangles. •In a parallelogram, opposite sides are equal. •In a parallelogram opposite angles are equal. •The diagonals of a parallelogram bisect each other. These properties have their converse also.
    • Special Parallelograms  If all the interior angles of a parallelogram are right angles, it is called a RECTANGLE. If all the sides of a parallelogram are congruent to each other, it is called a RHOMBUS.
    • If a parallelogram is both a rectangle and a rhombus, it is called a SQUARE. A Rhombus is a parallelogram with adjacent sides equal. The properties of rhombus are:- A rhombus has the including properties of A parallelogram. The diagonals of rhombus bisect each other at 90 degree The diagonals of rhombus bisect opposite angles
    • Properties of a Parallelograms When GIVEN a parallelogram, the definition and theorems are stated as ... A parallelogram is a quadrilateral with both pairs of opposite sides parallel. If a quadrilateral is a parallelogram, the 2 pairs of opposite sides are congruent.
    • If a quadrilateral is a parallelogram, the 2 pairs of opposite angles are congruent. If a quadrilateral is a parallelogram, the consecutive angles are supplementary. If a quadrilateral is a parallelogram, the diagonals bisect each other. If a quadrilateral is a parallelogram, the diagonals form two congruent triangles.
    • The line segment joining the mid point of two sides of a triangle is always parallel to the third side and half of it.
    • Given:-D and E are the mid points of the sides AB and AC . To prove:-DE is parallel to BC and DE is half of BC. construction:- Construct a line parallel to AB through C. proof:-in triangle ADE and triangle CFE AE=CE angle DAE= angle FCE (alternate angles ) angle AED= angle FEC (vertically opposite angles) Therefore triangle ADE is congruent to triangle CFE
    • Hence by CPCT AD= CF- - - - - - - - -1 But AD = BD(GIVEN) so from (1), we get, BD = CF BD is parallel to CF Therefore BDFC is a parallelogram That is:- DF is parallel to BC and DF= BC Since E is the mid point of DF DE= half of BC, and , DE is parallel to BC Hence proved .
    • Theorem : Sum of angles of a quadrilateral is 360 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. A B CD
    • THEOREM: The diagonal of a parallelogram divides it into two congruent triangles. Given: A parallelogram ABCD and its diagonal AC. To prove: Triangle ABC is congruent to triangle ADC Construction: 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. A B CD
    • THEOREM: In a parallelogram , opposite sides are equal. D A B C Given: A parallelogram ABCD. To Prove: AB = DC and AD = BC Construction: 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. 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=BC To 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 rule Since the triangles of a quadrilateral are equal, therefore it is a parallelogram.
    • B C THEOREM: In a parallelogram opposite angles are equal. A DGiven: A parallelogram ABCD. To prove: Angle A = Angle C & angle B=angle D Proof: 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. B 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 BC Similarly’ we can prove AB is parallel to CD. This shows that ABCD is a parallelogram.
    • THEOREM: The diagonals of a parallelogram bisect each other. B CD A O Given: A parallelogram ABCD To 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. B CD A O Given: In a quadrilateral ABCD, AO = OC & BO = OD To 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. B CD A Given: In a quadrilateral ABCD, AB is parallel to CD AB = CD To 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=1/2BC Proof: 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] C A B D E F
    • Now, AD= CF Also, AD = BD Therefore, CF = BD Again angle ADE = angle CFE (alternate angles) This implies that AD is parallel to FC Since, BD is parallel to CF (since, AD is parallel to CF and BD=AD). And, BD=CF Therefore, 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.
    • A B C D F 1 2 3 4 l M A E 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=CF Proof: 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.
    • Submitted by: Alma Buisan Submitted to: Ms. Charine Masilang