Here, AM is the median and AC and AM are perpendicular. The length BM comes out to be. BM=CM=2BC From the property of triangle. ∠A+∠B+∠C=180° In ΔAMC, ∠AMC+∠CAM+∠MCA=180° ∠AMC+90°+∠C=180° ∠AMC=90°−∠C Again at point M ∠BMP+∠PMA+∠AMC=180° ∠BMP+90°+90°−∠C=180° ∠BMP=∠C lnΔBPM ∠BPM+∠BMP+∠PBM=180° ∠BPM+∠C+∠B=180° Substituting the above values we get, ∠BPM=∠A Now, it is evident that ABC and MPB are equivalent triangle.
BM
BC
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BP
AB
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PM
AC
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Also,∠BAC=∠BAM+∠MAC ∠BAM=∠BAC−∠MAC ∠BAM=∠A−90° In ΔAMC ∠PAM+∠APM+∠AMP=180° ∠A−90°+∠APM+90°=180° tan(∠APM)=