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)
ျပင္ပ အမွတ္တစ္ခုမွ စက္၀ိုင္းတစ္ခုသို႔ secant တစ္ေၾကာင္းႏွင့္ tangent တစ္ေၾကာင္း ဆြဲေသာအခါ secant ၏ တစ္ေၾကာင္းလံုးႏွင့္ အျပင္ဘက္ပိုင္း တု႔ိ၏ အလ်ားမ်ားေျမွာက္လဒ္သည္ tangent ၏အလ်ား ႏွစ္ထပ္ကိန္းႏွင့္ ညီသည္။
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}$
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