Q. 95.0( 5 Votes )

# Two isosceles tri

Answer :

Given that ΔABC and ΔDBC are isosceles triangles.

Here, AB=AC and DB=DC

Also, ∠ABC=∠ACB

And ∠DBC=∠DCB

Now, in ΔABD and ΔACD,

AD=AD (common)

AB=AC (given)

BD=DC (given)

∴ ΔABD ≅ ΔACD (by SSS rule)

∠BAD=∠CAD (by cpct)

Again, in ΔABO and ΔACO,

AO=AO (common)

∠BAO=∠CAO (as ∠BAD=∠CAD)

AB=AC (given)

∴ ΔABD ≅ ΔACD (by SAS rule)

So, BO=OC (by cpct)

And ∠AOB=∠AOC(by cpct)

Also, ∠AOB+∠AOC=180° (straight angle)

⇒ ∠AOB+∠AOB=180°

⇒ 2∠AOB=180°

⇒ ∠AOB=90°

So, ∠AOC=∠AOB=90°

Hence, AD bisects BC perpendicularly.

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