in the given figure abcd is a quadrilateral in which ab=ad and be=de.Prove that the diagonals AC and BD bisect each other at right angles
Answers
Given: ABCD is a quadrilateral in which AB=AD and BC=DC
To prove: AC bisects ∠A and ∠C, and AC is the perpendicular bisector of BD
Proof:
In ∆ABC and ∆ADC, we have
AB = AD …given
BC = DC … given
AC = AC … common side
Thus by SSS property of congruence,
∆ABC ≅ ∆ADC
Hence, we know that, corresponding parts of the congruent triangles are equal
∠BAC = ∠DAC
∴ ∠BAO = ∠DAO …(1)
It means that AC bisects ∠BAD ie ∠A
Also, ∠BCA = ∠DCA … cpct
It means that AC bisects ∠BCD, ie ∠C
Now in ∆ABO and ∆ADO
AB = AD …given
∠BAO = ∠DAO … from 1
AO = AO … common side
Thus by SAS property of congruence,
∆ABO ≅ ∆ADO
Hence, we know that, corresponding parts of the congruent triangles are equal
∠BOA = ∠DAO
But ∠BOA + ∠DAO = 180°
2∠BOA = 180°
∴ ∠BOA = = 90°
Also ∆ABO ≅ ∆ADO
So, BO = OD
means that AC = BD
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