Hi, here is the question: I'm not exactly sure how this is solved, but I'm thinking similar triangles/congruency is involved as some smaller triangles share sides, perpendicular angles etc. Please help, BG
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$\displaystyle \widehat{MAQ}=90-\widehat{AMQ}$ $\displaystyle \widehat{PAK}=90-\widehat{APK}$ But PMQA is a cyclic quadrilater and $\displaystyle \widehat{AMQ}=\widehat{APK}$ Therefore, $\displaystyle \widehat{MAQ}=\widehat{PAK}$
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