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A B C D E is a regular pentagon. The bis...

`A B C D E` is a regular pentagon. The bisector of `/_A` of the pentagon meets the side `C D` in `M` Show that `/_A M C=90^@` In pentagon `ABCDE`

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In`A B C D E` pentagon
interior angle `=(5-2)xx180^0/5`
`=108^0`
in `ABCM` by angle sum
`/_B+/_C+/_BAM+/_AMC=360^0`
`108^0+108^0+54^0+/_AMC=360^0`
`/_AMC=90^0`
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