Question

please help me please yaar please ​

Answer 1

Answer:

$ABCDE \: is \: a \: regular \: pentagon.\\The \: bisector \: ∠A \: of \: the \: pentagon \: meets\\ the \: side \: CD \: at \: point \: M.\\ To \: prove : ∠AMC = 90° \\ Proof: \\ We \: know \: that, \: the measure \: of \\ each \: interior \: angle \: of \: a \: regular \: pentagon \: is \: 108° \\ ∠BAM = 1/2 × 108° = 54°\\ Since, \: we \: know \: that \: the \\sum \: of \: a \: quadrilateral \: is \: 360° \\ In \: quadrilateral \: ABCM, \: we \: have \\∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°\\ 54° + 108° + 108° + ∠AMC = 360° \\∠AMC = 360° - 270°\\∠AMC = 90°$

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