Question

AB= BC and AD=CD in given figure. Show BD bisects AC at
right angles.
no spam answers please​

Answer 1

Step-by-step explanation:

let there be angle(i),(ii),(iii)and (iv)

prove ∆ABE=∆BEC

you can prove it yourself

then....angle(i)=angle (iii)

how?

because its vertically opposite angles

same with angle(ii)=angle(iv)

when we take ∆ABE=∆BEC

we will find angle BEA=BEC

=90°

so as we did before angle AEB=angleDEC

and angleBEC=angleAED

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.