31st question plzzzzzzzzz solve it quickly
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Given : ABCD is a square
EF║BD
To prove : 1) DF = BE 2) AM bisects ∠BAD
Proof :
In Δ DCB
EF║BD
∴ E and F are the mid points. ( by converse of mod point theorem)
Now,
DC = CB
1/2 DC = 1/2 CB
DF = BE
2) Join MC
In Δ DCA and ΔCAB
AC = AC ( common)
DC = CB ( sides of square are
DA = AB equal)
∴Δ DCA congruent to Δ CAB ( By SSS)
∠DAM = ∠MAB ( by cpct)
∴ AM is the bisector.
EF║BD
To prove : 1) DF = BE 2) AM bisects ∠BAD
Proof :
In Δ DCB
EF║BD
∴ E and F are the mid points. ( by converse of mod point theorem)
Now,
DC = CB
1/2 DC = 1/2 CB
DF = BE
2) Join MC
In Δ DCA and ΔCAB
AC = AC ( common)
DC = CB ( sides of square are
DA = AB equal)
∴Δ DCA congruent to Δ CAB ( By SSS)
∠DAM = ∠MAB ( by cpct)
∴ AM is the bisector.
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