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ABCD is s rhombus. EABF is …

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ABCD is s rhombus. EABF is a straight line such that EA=AB=BF. Prove that ED and FC ,when produced ,meet at right angles.

Posted by Ramesh Tiwari 6 years ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 6 years ago

SinceΔEAD=ΔABC
(Corresponding angles),&AD=BC(sides of the rhombus)
By side angle side congrugency,the triangles
Therefore the corresponding parts of the triangle are also equal.
ΔCBF & ΔDAB
Since the sides
BF=AB
∠CBF∠DAB(corresponding angles)
ΔCBF&ΔDAB are congrugent
ΔAOB
∠ABO+∠BAO+∠AOB=180°
∠ABO+∠BAO=90°
∠CFB+∠AED=90°
∠XFE+∠XEF=90°

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CBSE > Class 09 > Mathematics