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Sagot :
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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