BE and CF are perpendicular on sides AC and AB of triangles ABC respectively . Prove that four points B , C ,E , F are concyclic. Also prove that the two angles of each of `DeltaAEFandDeltaABC` are equal.

Explanation

`BEbotAC, thereforeangleBEC=90^(@)`
Also `CFbotAB, thereforeanglesBFC=90^(@)` , i.e , at two points E and F the same side of BC , the produced angles are equal are equal.
`therefore` B ,C ,E ,F, are concyclic.
Now , in `DeltaAEF,angleAEF=angleAEB-angleBEF`
`90^(@)=angleBEF[becauseangleAEB=90^(@)]`
= `90^(@)-angleBCF [because angleBCF= angleBCF ` same angles in circle produced by the arc BF] ltbr `=90^(@)-angleBCF[becauseBEF=angleBCF` same angles in circle produced by the BF]
=`angleCBF[becauseangleBFC=90^(@)]`
Again , `angleAFE=angleAFC=angleCFE=90^(@)-angleCFE" " [becauseangleAFC=90^(@)]`
`=90^(@)-angleCBE[becauseangleCFE=angleCBE` , same angle in circle produced by the arc CE ]
`= angleBCE" " [becauseangleBEC=90^(@)]`
`therefore "in" Delta'sAEFand DeltaABC,angleAEF=angleABC" " [becauseangleCBF=angleABC]`
and `angleAFE=angleACB" " [becauseangleBCE=angleACB]`
Hence two angles of each of `DeltaAEFandDeltaACB` are equal.