Q14 of 56 Page 127

As shown in the figure of AB ⊥ BC, DC ⊥ BC and DE ⊥ AC then prove that ΔCED ~ ΔABC.

In ΔABC and ΔCED,


DC ⊥ BC


⇒ ∠ACB + ∠DCE = 90°


Let ∠ACB = x


⇒ ∠DCE = 90° – x


In ΔABC,


∠ACB + ∠ABC + ∠BAC = 180°


⇒ x + 90° + ∠BAC = 180°


⇒ ∠BAC = 90° – x


In ΔABC and ΔCED,


∠ABC = ∠DEC = 90°


∠ACB = ∠CDE = x


∠BAC = ∠DCE = 90° – x


Thus, the ΔABC and ΔCED are similar by AAA Similarity Rule.


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Answer the following in True or False. Write the reason of your answer (if possible).

(i) The ratio of the corresponding sides of two similar triangles is 4 : 9. Then the ratio of the areas of these triangles is 4 : 9.


(ii) In two triangles respectively ΔABC and ΔDEF of then ΔABC ≅ ΔDEF.


(iii) The ratio of the areas of two similar triangles in proportional to the squares of their sides.


(iv) If ΔABC and ΔAXY are similar and the values of their areas are the same then XY and BC may be coincident sides.