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#1

$Old$ October 23rd, 2009, 10:22 PM

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$Default$ proving isoceles tirangle

i need help with proving the 1st part.thx

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$Old$ October 23rd, 2009, 10:45 PM

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Originally Posted by helloying $View Post$

i need help with proving the 1st part.thx

HI

Since BC is parallel to AD , it is clear that $\angle CBD=\angle BDA$ and

$\angle BCA=\angle CAD$

Now use the properties of circles , where angles in the same segment are equal , so

$\angle BCA=\angle BDA$ and $\angle CBD=\angle CAD$ .

and you should be able to make your deduction now .

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