1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ΔABC ∠ ΔABD. What can you say about BC and BD ? In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD [Given] AB = AB [Common] ∠CAB = ∠DAB [∵ AB bisects ∠ CAD] ∴ Using SAS criteria, we have ΔABC ≌ ΔABD. ∵ Corresponding parts of congruent triangles (c.p.c.t) are equal. ∴ BC = BD. 2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that (i) ΔABD = ∠ ΔBAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC. (i) In quadrilateral ABCD, we have AD = BC and ∠DAB = ∠CBA. In ΔABD and ΔBAC, AD = BC [Given] AB = BA [Common] ∠DAB = ∠CBA [Given] ∴ Using SAS criteria, we have ΔABD ≌ ΔBAC (ii) ∵ ΔABD ≌ ΔBAC ∴ Their corresponding parts are equal. ⇒ BD = AC (iii) Since ΔABD ≌ ΔBAC ∴ Their corresponding parts are equal. ⇒ ∠ABD = ∠BAC. 3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. Solution: We have ∠ABC = 90° and ∠BAD = 90° Also AB and CD intersect at O. ∴ Vertically opposite angles are equal. Now, in ΔOBC and ΔOAD, we have ∠ABC = ∠BAD [each = 90°] BC = AD [Given] ∠BOC = ∠AOD [vertically opposite angles] ∴ Using ASA criteria, we have ΔOBC ≌ ΔOAD ⇒ OB = OA [c.p.c.t.] i.e. O is the mid-point of AB Thus, CD bisects AB. 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ∠ ΔCDA ∵ ℓ || m and AC is a transversal, ∴ ∠BAC = ∠DCA [Alternate interior angles] Also p || q and AC is a transversal, ∴ ∠BCA = ∠DAC [Alternate interior angles] Now, in ΔABC and ΔCDA, ∠BAC = ∠DCA [Proved] ∠BCA = ∠DAC [Proved] CA = AC [Common] ∴ Using ASA criteria, we have ΔABC ≌ ΔCDA 5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Figure). Show that: (i) ΔAPB ∠ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠ A. We have, ℓ as the bisector of QAP. ∴ ∠QAB = ∠PAB ∠Q= ∠P [each = 90°] ⇒ Third ∠ABQ = Third ∠ABP (i) Now, in ΔAPB and ΔAQB, we have AB = AB [common] ∠ABP = ∠ABQ [proved] ∠PAB = ∠QAB [proved] ∴ Using SAS criteria, we have ΔAPB ≌ ΔAQB (ii) Since ΔAPB ≌ ΔAQB ∴ Their corresponding angles are equal. ⇒ BP = BQ i.e. [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ] Thus, the point B is equidistant from the arms of ∠A. 6. In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE. We have ∠BAD = ∠EAC Adding ∠DAC on both sides, we have ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠DAE Now, in ΔABC and ΔADE, we have ∠BAC = ∠DAE [Proved] AB = AD [Given] AC = AE [Given] ∴ ΔABC ≌ ΔADE [Using SAS criteria] Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal. ⇒ BC = DE. 7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that (i) ΔDAP ∠ ΔEBP (ii) AD = BE We have, P is the mid-point of AB. ∴ AP = BP ∠EPA = ∠DPB [Given] Adding ∠EPD on both sides, we get ∠EPA + ∠EPD = ∠DPB + ∠EPD ⇒ APD = ∠BPE (i) Now, in ΔDAP ≌ ΔEBP, we have AP = BP. [Proved] ∠PAD = ∠PBE [∵ It is given that ∠BAD = ∠ABE] ∠DPA = ∠EPB [Proved] ∴ Using ASA criteria, we have ΔDAP ≌ΔEBP (ii) Since, ΔDAP ≌ΔEBP ∴ Their corresponding parts are equal. ⇒ AD = BE. 8. In 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 (see figure). Show that (i) ΔAMC ∠ ΔBMD (ii) ∠ DBC is a right angle. (iii) ΔDBC ∠ ΔACB (iv) CM = (1/2)AB Solution: ∵ M is the mid-point of AB. ∴ BM = AM [Given] (i) In ΔAMC and ΔBMD, we have CM = DM [Given] AM = BM [Proved] ∠AMC = ∠BMD [Vertically opposite angles] ∴ ΔAMC ≌ ΔBMD (SAS criteria) (ii) ∵ ΔAMC ≌ ΔBMD ∴ Their corresponding parts are equal. ⇒ ∠MAC = ∠MBD But they form a pair of alternate interior angles. ∴ AC || DB Now, BC is a transversal which intersecting parallel lines AC and DB, ∴ ∠BCA + ∠DBC = 180° But ∠BCA = 90 [∵ ΔABC is right angled at C] ∴ 90° + ∠DBC = 180° or ∠DBC = 180° ∠ 90° = 90° Thus, ∠DBC = 90° (iii) Again, ΔAMC ≌ ΔBMD [Proved] ∴ AC = BD [c.p.c.t] Now, in ΔDBC and ΔACB, we have ∠DBC = ∠ACB [Each = 90°] BD = CA [Proved] BC = CB [Common] ∴ Using SAS criteria, we have ΔDBC ≌ ΔACB (iv) ∵ ΔDBC ≌ ΔACB ∴ Their corresponding parts are equal. ⇒ DC = AB But DM = CM [Given] ∴ CM = (1/2) DC = (1/2) AB ⇒ CM = (1/2) AB