Q9 of 13 Page 96

Two isosceles triangles ABC and DBC whose AB = AC, DB = DC and they are situated on the opposite side of BC. Let’s prove that AD bisects BC perpendicularly.


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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