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In traangle ABC , AB =AC and E is any point on the extended BC . If the circumcircle ofDeltaABC intersect AE at the point D then that angleACD=angleAEC. |
Answer» Solution :In `DeltaABC`, AB=AC, `thereforeangleABC=angleACB` ………….(1) Now , ABCD is a CYCLIC QUADRILATERAL, `thereforeangleABC+angleADC=180^(@)` or , `angleACB+angleADC=180^(@)` ……..(2) [from (1)] , Again , `angleACB+angleACE=1" straight angles " =180^(@)` ........(3) From (2) and (3) we GET , `angleACB+angleADC=angleACB+angleACE` or, `angleADC=angleACE" or " angleDCE+angleDEC=angleACD+angleDCE" or " angleDEC=angleACD` `thereforeangleACD=angleDEC` Hence `angleACD=angleAEC` ![]() |
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