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. Prove that, AD is the

perpendicular bisector of BC.​

Answer 1

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