Geometry

Hong Kong

Stage 1 - Stage 3

$ABCD$`A``B``C``D` is a parallelogram with $AE=FC$`A``E`=`F``C`. Prove that $DE$`D``E`=$FB$`F``B`.

In $\triangle AED$△`A``E``D` and $\triangle CFB$△`C``F``B` we have :

Easy

Approx 6 minutes

$ABCD$`A``B``C``D` is a parallelogram, with $AY$`A``Y`=$XC$`X``C`.

Prove that the two triangles $\triangle AYB$△`A``Y``B` and $\triangle CXD$△`C``X``D` are congruent.

In the diagram, $ABCD$`A``B``C``D` is a square, and $AE=CF$`A``E`=`C``F`. Prove that $AF=EC$`A``F`=`E``C`.

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