I need ans for question 8
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(a) In ΔADC and ABC
∠DAC=∠BAC(given)
∠DCA=∠BCA(given)
AC = AC(common)
∴ΔADC = ΔABC
therefore by CPCT , AD= AB and BC = DC
⇒ABCD is a square
(b)
In ΔBAD and BCD
BD = BD
AD = DC
BC = AB
by SSS rule ΔBAD =Δ BCD
∴∠CDB = ∠BDA
∠ABD=∠CBD
∴BD bisects ∠B and ∠C
PROVED.........
∠DAC=∠BAC(given)
∠DCA=∠BCA(given)
AC = AC(common)
∴ΔADC = ΔABC
therefore by CPCT , AD= AB and BC = DC
⇒ABCD is a square
(b)
In ΔBAD and BCD
BD = BD
AD = DC
BC = AB
by SSS rule ΔBAD =Δ BCD
∴∠CDB = ∠BDA
∠ABD=∠CBD
∴BD bisects ∠B and ∠C
PROVED.........
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hope this helps
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