Q. 14

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


Answer :

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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