Question

solve this question step by step plzzz​

Answer 1

in this ΔAPO and ΔBPO

AP=BP(tangents from an external pt. is equal)

PO=PO(common to both)

angle A°= angle B°(as AP=BP so angle A=angle B)

so ΔAPO is congruent to ΔBPO

so angle APO=BPO(CPCTE)

but they are the same thing and their addition will give full angle P

so PO is the bisector of angle P

so ΔAPB is an equilatral triangle