In the eighth grade mathematics equilateral triangle ABC, there is a point P, AP is equal to 3, BP is equal to 1, CP is equal to 5?
As shown in the figure, take AP as the edge, make equilateral △ APD, and connect BD
Then ∠ bad = 60 ° - ∠ BAP = ∠ cap,
In △ ADB and △ APC,
AD=AP.∠BAD=∠CAP,AB=AC
∴△ADB≌△APC(SAS)
The results show that BD = PC = 5, PD = AP = 3, BP = 4
∴BP2+PD2=42+32=25=BD2
∴∠BPD=90°
∴∠APB=∠APD+∠BPD=150°.
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