InterviewSolution
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InterviewSolution
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ABC is a cyclic equailateral triangle. If D be any point on the circular are BC on the oppsite side of the point A, then prove that DA=DB+DC |
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Answer» Solution :Let ABC be a cyclic equailateral angle inside the circle with centre at O,D is any point on the circular are BC on the opposite side of thepoint A. We have to prove that`DA=DB+DC`  ConstructionLet us cut a part De from DA equal to DC. i.e, DC= DEand let us join C,E In `Delta CDE, DC= DE ` [ as per construction] `:. angle DCE=angle DEC.....(1)` `:. Delta ABC` is an equilateral triangle, `angleBAC=angle ACB=angle CBA=60^(@) and AB BC= CA ......(2)` Now, TWO angles in circle produced by the are AC are `angle ADC and angle ABC` `:. angle ADC= 60^(@)`[ by (2)] `:. angle CDE= 60^(@)......(3)` `:."in" Delta CDE, angle CDE+ angle DCE= 180^(@)+ angle DEC=180^(@)` or, `60^(@)+angleDCE+ angle DCE= 180^(@)`[ from (1) and (3)] or, ` 2angle DCE =180^(@)-60^(@) or, angle DCE=(120^(@))/(2)=60^(@)` `:. angle DEC=60^(@)` i.,e, each and very angle of `Delta CDE` is `60^(@), Delta CDE` is equilateral. `:. CD=DE=CE....(4)` Now, `angle ACE+angleBCE= angleACB=60^(@)= angle DCE= angle BCD+angle BCE` `:. angle ACE= angle BCD........(5)` Againin `Delta`'s ACE and `Delta BCD` `angle CAE= angle CBD[ :.` bothare angles in cricle produced by the same chord CD] `:. angle ACE= angle BCD` [by (5)] andAC=BC [ `:. Delta ABC` is equilateral] `:. Delta ACE~= BCD [ :.` by the A-A-S condition of congruency] `:. AE=BD [ :.` similar sides of congruent tirangles]........(6) THEREFORE, `DA=DE+AE=DC+BD [ :. DE= DC and AE= BD]` Hence `DA=DB+DC` |
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