If a secant and a tangent aredrawn to a circle from an external point the square the of the tangent segment is equal to the product of the length of the secant segment and its external part. (Theorem-6 from grade 11 Mathematics TextBook)
ျαα္α α‘αွα္αα
္αုαွ α
α္αိုα္းαα
္αုαိုα secant αα
္ေαΎαာα္းႏွα့္ tangent αα
္ေαΎαာα္း αြဲေαာα‘αါ secant ၏ αα
္ေαΎαာα္းαံုးႏွα့္ α‘ျαα္αα္αိုα္း αုαိ၏ α‘α်ားα်ားေျαွာα္αα္αα္ tangent ၏α‘α်ား ႏွα
္αα္αိα္းႏွα့္ αီαα္။
CE touches the circle BAED at E and circle CAB at C and DF touches the circle CAB at F. If CAD is a straight line, prove that CE² + DF² = CD².
$ \displaystyle \begin{array}{l}\text{Proof : In smaller circle, }CAFB\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ C{{E}^{\text{2}}}=CA\cdot CD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{In larger circle, }BAED\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ D{{F}^{\text{2}}}=CD\cdot AD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore C{{E}^{\text{2}}}+D{{F}^{\text{2}}}=CA\cdot CD+CD\cdot AD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =CD\left( {CA+AD} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =CD\cdot CD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore C{{E}^{\text{2}}}+D{{F}^{\text{2}}}=C{{D}^{\text{2}}}\end{array}$
α
ာαα်αူ၏ α‘αြα်αို αေးα
ားα
ွာα
ောα့်αျှော်αျα်!
Thanks.
ReplyDeletePost a Comment