Problem — The diagonals of an isosceles trapezoid are congruent

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Let $EFGH$ be an isosceles trapezoid with bases $EF$ and $HG$ (where $EF \parallel HG$) and congruent legs $EH \cong FG$. Prove that its diagonals are congruent, that is $EG \cong FH$.

(original figure: isosceles trapezoid $EFGH$ with the two diagonals)

Solution

Base angles. Through $H$ draw the parallel to the leg $FG$; it meets the longer base $EF$ at a point $K$. The quadrilateral $KFGH$ has $KF \parallel HG$ and $HK \parallel FG$, hence it is a parallelogram and $HK \cong FG$. But $FG \cong EH$ by hypothesis, so $HK \cong EH$: the triangle $EHK$ is isosceles and therefore $\widehat{HEK} \cong \widehat{HKE}.$ Moreover $\widehat{HKE} \cong \widehat{GFE}$, being corresponding angles of the parallel lines $HK$ and $FG$ cut by the transversal $EF$. Hence the angles at the longer base are congruent: $\widehat{HEF} \cong \widehat{GFE}.$ Congruence of the diagonals. Compare the triangles $HEF$ and $GFE$:

• $EF$ is common;
• $EH \cong FG$ (hypothesis);
• $\widehat{HEF} \cong \widehat{GFE}$ (base angles, just proved).

By the SAS congruence criterion $HEF \cong GFE$, hence the corresponding sides $HF$ and $GE$ are congruent. That is, the diagonals $FH \cong EG$. $\;\blacksquare$