# Proof: By Euclid

• Let the parallelogram $BEFG$, equal to the triangle $C$, have been constructed in the angle $EBG$, which is equal to $D$ [Prop. 1.42].
• And let it have been placed so that $BE$ is straight-on to $AB$.1
• And let $FG$ have been drawn through to $H$, and let $AH$ have been drawn through A parallel to either of $BG$ or $EF$ [Prop. 1.31], and let $HB$ have been joined.
• And since the straight line $HF$ falls across the parallels $AH$ and $EF$, the (sum of the) angles $AHF$ and $HFE$ is thus equal to two right angles [Prop. 1.29].
• Thus, (the sum of) $BHG$ and $GFE$ is less than two right angles.
• And (straight lines) produced to infinity from (internal angles whose sum is) less than two right angles meet together [Post. 5] .
• Thus, being produced, $HB$ and $FE$ will meet together.
• Let them have been produced, and let them meet together at $K$.
• And let $KL$ have been drawn through point $K$ parallel to either of $EA$ or $FH$ [Prop. 1.31].
• And let $HA$ and $GB$ have been produced to points $L$ and $M$ (respectively).
• Thus, $HLKF$ is a parallelogram, and $HK$ its diagonal.
• And $AG$ and $ME$ (are) parallelograms, and $LB$ and $BF$ the so-called complements, about $HK$.
• Thus, $LB$ is equal to $BF$ [Prop. 1.43].
• But, $BF$ is equal to triangle $C$.
• Thus, $LB$ is also equal to $C$.
• Also, since angle $GBE$ is equal to $ABM$ [Prop. 1.15], but $GBE$ is equal to $D$, $ABM$ is thus also equal to angle $D$.
• Thus, the parallelogram $LB$, equal to the given triangle $C$, has been applied to the given straight line $AB$ in the angle $ABM$, which is equal to $D$.
• (Which is) the very thing it was required to do.

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