Q4 of 65 Page 150

ABC is an isosceles triangle right angled at C. Prove that

Given: ΔABC is an isosceles triangle

AC = CB


Using Pythagoras theorem in ΔABC (i.e., right-angled at point C), we get


AC2 + CB2 = AB2


AC2 + AC2 = AB2 (AC = CB)


2AC2 = AB2


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In Fig. 6.53, ABD is a triangle right angled at Aand AC ⊥ BD. Show that:

(i) AB2 = BC . BD


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(iii) AD2 = BD . CD


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