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Answer:

If three sides of a triangle are equal and the measure of all three angles is equal to 60 degrees then the triangle is an equilateral triangle.  Therefore it is an equilateral triangle.

In ∆PAM and ∆PBM

[tex]\because\pink{\sf\begin{cases} \sf AM=BM(Bisected\:by\:PM)\\ \sf PA=PB(Given)\\ \sf PM(Common\:side)\end{cases}}[/tex]

[tex]\\ \sf\longmapsto ∆PAM\cong ∆PBM(SSS)[/tex]

[tex]\\ \sf\longmapsto PA+AM=PB+BM[/tex]

  • Put AM =BM

[tex]\\ \sf\longmapsto PA+AM+AM=PB+BM+BM[/tex]

  • Put PA=PB

[tex]\\ \sf\longmapsto PA+AB=PB+AB[/tex]

Hence

∆APB is a equilateral triangle

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