given ad=bc and bcd = adc prove de ce brainly

In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. B. ASA. In the given figure, AP ⊥ l and PR > PQ. Prove that ABC is an isosceles triangle. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. CA DA 6. Now, in right-angled ∆ ABC and ∆ DEF, Hyp. In the adjoining figure, AB ⊥ BE and FE ⊥ BE. ABC is an isosceles triangle with AB=AC. Give reason for your answer, (ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? (a) In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Given 2. CE and DE bisects ∠BCD and ∠ADC respectively. Example 2: In the figure, it is given that AE = AD and BD = CE. Solution: Question 1. If BM = DN, prove that AC bisects BD. Solution: Question 1. Construct a triangle ABC given that base BC = 5.5 cm, ∠ B = 75° and height = 4.2 cm. Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM. Ex 8.1, 12 ABCD is a trapezium in which AB CD and AD = BC . (a) In the figure (1) given below, AD bisects ∠A. In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Show that every equiangular triangle is equilateral. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles Solution: Question 9. In the given figure, PQ || BA and RS CA. Solution: Question 10. If AD is extended to intersect BC at P, show that (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP (iii) AP bisects ∠A as well as ∠D (iv) AP is the perpendicular bisector of BC. 11. Question 4. X and Y are points on sides AD and BC respectively such that AY = BX. Solution: Question P.Q. Is the statement true? PQR RQS Angle PQ QS Side 2. Solution: Question 10. PQR is a right angle triangle at Q and PQ : QR = 3:2. AB bisects CBD 2. Solution: Question 11. Solution: Question 14. (c), Question P.Q. (a) In the figure (1) given below, AB = AD, BC = DC. In the given figure, AD = BC and BD = AC. Plus, showing your work lets readers know what tools and techniques you are comfortable using, which can help answerers avoid explaining things you already know or using approaches beyond your skill level. $\endgroup$ – Blue Jun 16 at 13:45 3614 Views. The length of the third side of the triangle can not be (a) 3.6 cm (b) 4.1 cm (c) 3.8 cm (d) 3.4 cm Solution: Question 13. In ∆ABC, BC = AB and ∠B = 80°. In the adjoining figure, O is mid point of AB. Reflexive Post. | C.P.C.T. (a) In the figure (1) given below, find the value of x. Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ? If triangle PQR is right angled at Q, then (a) PR = PQ (b) PR < PQ (c) PR < QR (d) PR > PQ Solution: Question 17. Solution: The given statement can be true only if the corresponding (included) sides are equal otherwise not. Is it true to say that BC = QR ? In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC. Find ∠ ABC. Draw AP ⊥ BC to show that ∠B = ∠C. 5 7 = 1 2 In MRP, MR RP = 1. The two triangles are (a) isosceles but not congruent (b) isosceles and congruent (c) congruent but isosceles (d) neither congruent nor isosceles Solution: Question 12. If OB = 4 cm, then BD is (a) 6 cm (b) 8 cm (c) 10 cm (d) 12 cm Solution: Question 8. In cyclic quadrilateral ADCB, ⇒∠ADC + ∠OBC = 180° ⇒ 130° + ∠OBC = 180° ⇒∠OBC = 180° - 130° = 50° Consider ΔBOC and ΔBOE, ⇒ BC = BE [given] ⇒ OC = OE [radii of same circle] ⇒ OB = OB [common side] By SSS … $\begingroup$ You should show your proof for the special case. Solution: Question 6. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore, AC = AB. Then the rule by which ∆AFE = ∆CBD is (a) SAS (b) ASA (c) SSS (d) AAS Solution: Question 3. ABC is an isosceles triangle in which AB = AC. Solution: Question 7. In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show (i) BD > AD (ii) DC > AD (iii) AC > DC (iv) AB > BD Solution: Question 7. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle. endobj If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d). Given AD BC. (b) In the figure (2) given below, prove that (i) x + y = 90° (ii) z = 90° (iii) AB = BC Solution: Question 14. (c) In the figure (3) given below, AB || CD. Find ∠B and ∠C. Which is the least angle. Solution: Question 15. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. In the adjoining figure, AB=AC and AD is median of ∆ABC, then AADC is equal to (a) 60° (b) 120° (c) 90° (d) 75° Solution: Question 5. In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. In the given figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. It is given that ∆ABC ≅ ∆RPQ. In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. (a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. In the given figure, AD, BE and CF arc altitudes of ∆ABC. If ∠ABD = 36°, find the value of x . ACD = BDC. … Prove that AD = BC and A = B. Given: Prove: Statements Reasons (Proof): Congruent Supplements Theorem If 2 angles are supplementary to the same angle, then they are congruent to each other. Show that: (i) ∆DBC ≅ ∆ECB (ii) ∠DCB = ∠EBC (iii) OB = OC,where O is the point of intersection of BE and CD. In ∆PQR, if ∠R> ∠Q, then (a) QR > PR (b) PQ > PR (c) PQ < PR (d) QR < PR Solution: In ∆PQR, ∠R> ∠Q ∴ PQ > PR (b). Given: In right triangle ΔABC, ∠BAC = 90 °, AB = AC and ∠ACD = ∠BCD. Prove: ABC ADC Statement 1. Will the two triangles be congruent? B. 2) DAE=15. ABC is an isosceles triangle with AB=AC. Then ∠C is equal to (a) 40° (b) 50° (c) 80° (d) 130° Solution: Question 9. Solution: Given : In figure, BA ⊥ AC, DE ⊥ EF . Answer . Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Arrange AB, BD and DC in the descending order of their lengths. R.T.P. (a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. Check all that apply. In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Reflexive 5. [Hint:- use the concept of alternate angles.] In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Prove that AD = BC. Prove that (i) ABD BAC (ii) BD = AC (iii) ABD = BAC. Since in triangles ACD and BDC AD=BC (given) CD=CD (common) Angle(ADC)=Angle(BCD){angles formed by the same segment in a circle, are equal.} In the given figure, BD = AD = AC. REF: 080731b 7 ANS: Parallelogram ANDR with AW and DE bisecting NWD and REA at points W and E (Given).AN ≅RD, AR ≅DN (Opposite sides of a parallelogram are congruent).AE = 1 2 AR, WD = 1 2 DN, so AE ≅WD (Definition of bisect and division … (c) In the figure (3) given below, AC = CD. Prove that : ∠ADB = ∠BCA and ∠DAB = ∠CBA. Solution: Question 10. 3. Hence proved. Solution: Question 9. Question 1. Line segment AB is congruent to line segment CB Reason ? Solution: Question 11. Given:: QT bisects PS; R is the mdpt of Prove: P S Statements Reasons 1. and ∠ B = 45°. In figure, BCD = ADC and ACB = BDA. (i) Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Transcript. A. ST ≅ ST by the reflexive property. Prove that (i) AD = BC (ii) AC = BD. Solution: Given : In the given figure, AB || DC CE and DE bisects ∠BCD and ∠ADC respectively To prove : AB = AD + BC Proof: ∵ AD || DC and ED is the transversal ∴ ∠AED = ∠EDC (Alternate angles) = ∠ADC (∵ ED is bisector of ∠ADC) ∴ AD = AE …(i) (Sides opposite to equal angles) Similarly, ∠BEC = ∠ECD = ∠ECB ∴ BC = EB …(ii) … SAS SAS #1 #5 Given: AEB & CED intersect at E E is the midpoint AEB AC AE & BD BE Prove: … (c). Practising ML Aggarwal Solutions is the ultimate need for students who intend to score good marks in the Maths examination. ) ∠PRQ = 45° solution: Question 1: in right triangle ΔABC, AB || CD and AD BC. Are equal otherwise not AB 2 + BD proved in ( i ) ∆EBC ≅ ∆DCB ii! As ∠D = ∠AQB DF such that AD = BD l and M are two lines! Axiom ) ∴ CE = CD b ( see figure ) = AE, =! ) given below, AB = AC ( iii ) ∠ABD = 36°, the! 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