T H E E L E M E N T S |
Book I Proposition 5 | |
And, since AF is equal to AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG. But FC was also proved equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base is common to them; therefore the triangle BFC is equal to the triangle CGB, and the remaining angles will be equal to rhe remaining angles respectively, namely those which the equal sides subtend; thus the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. |
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