In the diagram ABCD is a trapezium with AB CD Prove that i BAD is congruent to

# In the diagram abcd is a trapezium with ab cd prove

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4.In the diagram, ABCDis a trapezium with AB = CDProve that (i) BADis congruent to CDA(ii) . . , , s) ) BC ) ) .
24 5. In the diagram, BCDE is a rectangle. AED is isosceles with AD = AE . AFB and AGC are straight lines. Show that (i) ABE is congruent to ACD , (ii) AEF is congruent to ADG . (Pg256Q13) Proof (i) AED = ADE (base s of isos. AED ) BEF = CDG = 90 ( s of rectangle) AED + BEF = ADE + CDG Hence, AEB = ADC . AE = AD (given) BE = CD (opp sides of rectangle) Hence ABE is congruent to ACD . (SAS) (ii) EAF = DAG (corr s of congruent s) AE = AD (given) AEF = ADG (base s of isos ) Hence AEF is congruent to ADG . (ASA) 6. In the diagram, ABCD and, CDEF are parallelograms. The point X is the mid-point of AC and of BD , and the point Y is the mid-point of CE and of DF . Prove that (i) AE is parallel to XY , (ii) AE is parallel to BF , (iii) BAEF is a parallelogram. (Pg256Q14) Proof (i) AE // XY (Mid-point Theorem) (ii) XY // BF (Mid-point Theorem) Hence, AE // XY // BF . (iii) AB // CD and CD // EF Hence, AB // EF . BAEF is a parallelogram. sgexamguru.com
25 7. In the diagram below, ABCD is a quadrilateral. AE = BE and BF = FC . The diagonal AC intersects DE at X and DF at Y , so that AX = XY = YC . Prove that (i) BXDY is a parallelogram, (ii) AXB is similar to CYD . (iii) ABCD is a parallelogram. (Pg270Q5) Proof (i) EX // BY (Mid-point Theorem) FY // BX (Mid-point Theorem) Hence, BXDY is a parallelogram. (ii) AXB = XYF (corr. s, BX // FY ) = CYD (vert. opp. s) AX = CY (given) BX = DY (opp. sides of parallelogram) AXB is congruent to CYD . (SAS) (iii) BAX = DCY . (corr s of congruent s) AB // CD (alt s are equal) AB = CD (corr sides of congruent s) ABCD is a parallelogram. sgexamguru.com
26 Singapore Chinese Girls’ School Secondary 4 Additional Mathematics Plane Geometry Name: ( ) Class: Sec 4_____ Date: ________________ Worksheet 7: Angle Properties of Circles Some Angle Properties of Circles Property Abbreviation 1. An angle in a semicircle is a right angle. in a semicircle 2. An angle at the centre is twice any angle at the circumference. at centre = 2 at ce 3. Angles in the same segment are equal. s in the same segment 4. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Ext of a cyclic quad 5. Angles in opposite segments are supplementary. x + y = 180 s in opp segments (or opp s of a cyclic quad) sgexamguru.com
27 Property Abbreviation 6. A tangent of a circle is perpendicular to the radius drawn to the point of tendency. OAB = 90 tan rad 7. If PQ and PR are two tangents to a circle centred at O , then PQ = PR OPQ = OPR tan from an ext pt The Alternate Segment Theorem (Tangent-Chord Theorem) An angle between a tangent, ATB and a chord, TP through the point of contact, T , is equal to the angle in the alternate segment.