Question

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

Answer 1

Answer:

In ∆BAO and ∆DAO

BO=DO

at O 90°( right angle)

AO=AO common side

∆BAO =~ ∆DAO

By SAS congruent

BA=AD {CPCT/ hypotenuse}

In ∆BOC and ∆ DOC

BO=DO

at O 90°

CO=CO by common side

∆BOC=~∆ DOC

by SAS congruent

therefore

BC=DC (Hypotenuse/cpct)

I hope it's helpful to you

Answer 2

Question-

In the following figure, AB = BC and AD = CD. Show that BD bisects AC at right angles.

Answer-

We are required to prove ∠BEA = ∠BEC = 90° and AE = EC.Consider ∆ABD and ∆CBD,

AB = BC (Given)

AD = CD (Given)

BD = BD (Common)

Therefore, ∆ABD ≅ ∆CBD (By SSS congruency)

∠ABD = ∠CBD (CPCT)

Now, consider ∆ABE and ∆CBE,

AB = BC (Given)

∠ABD = ∠CBD (Proved above)

BE = BE  (Common)

Therefore, ∆ABE≅ ∆CBE (By SAS congruency)

∠BEA = ∠BEC (CPCT)

And ∠BEA +∠BEC = 180° (Linear pair)

2∠BEA = 180° (∠BEA = ∠BEC)

∠BEA = 180°/2 = 90° = ∠BEC

AE = EC (CPCT)

Hence, BD is a perpendicular bisector of AC.