In the figure $∠$ ABC = $∠$ CDE = 90 $°$ seg AC $≅$ seg CE seg BC $≅$ seg ED
show that:
(i) $△$ ABC $≅$ $△$ CDE
(ii) $∠$ BAC $≅$ $∠$ ECD
(iii) $∠$ ACE = 90 $°$

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Solution

$Given : ∠ ABC = ∠ CDE = 90 ° seg AC ≅ seg CE seg BC ≅ seg ED (i) In △ ABC and △ CDE, ∠ ABC = ∠ CDE = 90 ° (given) seg AC ≅ seg CE (given) seg BC ≅ seg ED (given) Thus, △ ABC ≅ △ CDE (h ypotenuse - s ide theorem) (ii) By c . a . ct, ∠ BAC = ∠ ECD$

$(iii) Since △ ABC ≅ △ CDE, ∠ BAC = ∠ ECD (by c . a . c . t) . . . (1) and ∠ ACB = ∠ CED (By c . a . c . t) . . . (2) In △ ABC, ∠ BAC + ∠ ABC + ∠ ACB = 180 ° (a ngle sum property) \Rightarrow ∠ BAC + 90 ° + ∠ ACB = 180 ° \Rightarrow ∠ BAC + ∠ ACB = 90 ° \Rightarrow ∠ ECD + ∠ ACB = 90 ° \{From (1)\} . . . . . . (3) Since BCD is a straight line, ∠ ACB + ∠ ACE + ∠ ECD = 180 ° \Rightarrow ∠ ACE + 90 ° = 180 ° \{f rom (3)\} \Rightarrow ∠ ACE = 90 ° Hence p roved .$

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