Question

IN A TRIANGLE , ANGLE A=60°. PROVE THAT BC²=AB²₊AC²₋AB·AC

Answer 1

Construction-draw cd perpendicular ab
In right triangle BCD
BC*2=CD*2+BD*2
In right triangle BCD
bc*2=dc*2+dc*2
bc*2=DC*2+(ab-ad)*2
BC*2=DC*2+ab*2+ad*2-2ab.ad
Now DC*2=ac*2-ad*2
BC*2=ac*2+ab*2-2ab.ad
Now in triangle ADC angle ACD=30°
Therefore, da=1\2ac. {Prop.of 30°,60°&90°}
BC*2=ac*2+ab*2-2ab.1\2ac
BC*2=ac*2+ab*2-ab.ac