Determine int(a) and cl(a) in each case. (a) a = (0, 1] in the lower limit topology on r. (b) a = {a} in x = {a, b, c} with topology {x, ø, {a}, {a, b}}. (c) a = {a, c} in x = {a, b, c} with topology {x, ø, {a}, {a, b}}. (d) a = {b} in x = {a, b, c} with topology {x, ø, {a}, {a, b}}. (e) a = (-1, 1) u {2} in the standard topology on r. (f) a = (-1, 1) u {2} in the lower limit topology on r. (g) a = {(x, 0) erp | xer} in r2 with the standard topology. (h) a = {(0, x) er| xer} in r2 with the topology generated by the basis in exercise 1.19. (i) a = {(x, 0) erp | xer} in r2 with the topology generated by the basis in exercise 1.19.

Given: ad¯¯¯¯¯ is an altitude. prove: ab2+ac2=cb2 right triangle a b c with right angle a. point d lies on side b c and segment a d is drawn. angle a d c is a right angle. drag and drop a reason into each box to correctly complete the two-column proof. statement reason ad¯¯¯¯¯ is an altitude, and ∠bac is a right angle. given ∠adb and ∠adc are right angles. definition of altitude ∠bac≅∠bda ? ∠bac≅∠adc ? ∠b≅∠b ? ∠c≅∠c reflexive property of congruence △abc∼△dba ? △abc∼△dac aa similarity postulate abbd=cbab ? ab2=(cb)(bd) cross multiply and simplify. acdc=cbac polygon similarity postulate ac2=(cb)(dc) cross multiply and simplify. ab2+ac2=ab2+(cb)(dc) addition property of equality ab2+ac2=(cb)(bd)+(cb)(dc) substitution property of equality ab2+ac2=(cb)(bd+dc) ? bd+dc=cb segment addition postulate ab2+ac2=cb2 substitution property of equality