Question

in the given figure, AC bisects BD at right angle. prove that AB and BC=DC​

Answer 1

Step-by-step explanation:

abhrajit76

08.07.2019

Math

Secondary School

+5 pts

Answered

in the fig ABCD is a quadrilateral in which AB=AD and BC=DC.. prove that 1. AC bisects angles A and C

2

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shahbazalam987p65ymd

shahbazalam987p65ymd

Answer :-

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

Which means that AC = BD

Answer 2

Answer:

ANSWER:

Given: ABCD is a quadrilateral in which AB = AD and BC = DC

(i)

In ∆ABC and ∆ADC, we have:

AB = AD

(Given)

BC = DC

(Given)

AC is common.

i.e., ∆ABC ≅ ∆ADC

(SSS congruence rule)

∴ ∠BAC = ∠DAC and ∠BCA = ∠D CA

(By CPCT)

Thus, AC bisects ∠A and ∠ C.

(ii)

Now, in ∆ABE and ∆ADE , we have:

AB = AD

(Given)

∠BAE = ∠DAE

(Proven above)

AE is common.

∴ ∆ABE ≅ ∆ADE

(SAS congruence rule)

⇒ BE = DE

(By CPCT)

(iii) ∆ABC ≅ ∆ADC

∴ ∠ABC = ∠AD C

(Proven above)

(By CPCT)