Hello, I have a small problem, so I decide to ask You for a help. I don't know how improve that area AEF is equal to area of ABE and ADF triangels (together) Thank You for Your time Have a nice day
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Originally Posted by rudzielica Hello, I have a small problem, so I decide to ask You for a help. I don't know how improve that area AEF is equal to area of ABE and ADF triangels (together) Thank You for Your time Have a nice day you mean 'to prove' not 'improve' Let $H$ be the point symmetric of $D$ with respect to $AF$ Area of triangle $AFH=$ area $AFD$ similarly construct a triangle $AEK$ whose area is equal to the area of triangle $AEB$
Last edited by Idea; Sep 21st 2019 at 11:58 AM.
Thank You for a help and correction. However, I don't know where is point K? And there isn't informationa that DF+BE=FE
$K$ is symmetric of $B$ w.r.t. the line $AE$ $AFH$ and $AEK$ are right triangles angle $AHF=90$ degrees angle $AKE=90$ degrees Show that area $AFH+AEK=AFE$ Note that $A,H,K$ are collinear