Question

ABC and DBC are two triangles on the same
base BC such that A and D lie on the opposite
sides of BC, AB AC and DB = DC. Show that AD
is the perpendicular bisector of BC.
​

Answer 1

Step-by-step explanation:

ANSWER

Let AD intersect BC at O

Then we have to prove that ∠AOB=∠AOC=90

∘

and BO=OC

In △ABD and △ACD we have

AB=AC(given)

AD=DA(common)

BD=DC(given)

∴△ABD≅△ACD

⇒∠1=∠2 proved above

∴∠AOB=∠AOC

But ∠AOB+∠AOC=180

∘

(linear pair)

⇒∠AOB+∠AOB=180

∘

⇒2∠AOB=180

∘

∴∠AOB=90

∘

Hence AD is perpendicular to BC and AD bisects BC

∴AD is the perpendicular bisector of BC

Answer 2

Step-by-step explanation:

ANSWER

Let AD intersect BC at O

Then we have to prove that ∠AOB=∠AOC=90

∘

and BO=OC

In △ABD and △ACD we have

AB=AC(given)

AD=DA(common)

BD=DC(given)

∴△ABD≅△ACD

⇒∠1=∠2 proved above

∴∠AOB=∠AOC

But ∠AOB+∠AOC=180

∘

(linear pair)

⇒∠AOB+∠AOB=180

∘

⇒2∠AOB=180

∘

∴∠AOB=90

∘

Hence AD is perpendicular to BC and AD bisects BC

∴AD is the perpendicular bisector of BC