Question

In a triangle ABC, AB = AC, D and E are points on the sides AB and AC, respectively such that BD = CE. Show that:

1. ΔDBC ≅ ΔЕСВ
2. ΔDCB = ΔЕВС
3. OB = OC, where O is the point of intersection of BE and CD.

Answer 1

We are given:

• Triangle ABC, AB = AC (i.e., it is isosceles)
• Points D and E lie on sides AB and AC respectively
• BD = CE

We are to show:

i. ΔDBC ≅ ΔЕСВ

In triangles DBC and ECB:

• BD = CE   ...(Given)
• AB = AC   ...(Given ⇒ so ∠ABC = ∠ACB since triangle ABC is isosceles)
• BC = BC   ...(Common side)

So, in these triangles:

• BD = CE   ...(Given)
• BC = BC   ...(Common)
• ∠DBC = ∠ECB   ...(These are the same angles from the isosceles triangle)

So by SAS congruence, we get:

ΔDBC ≅ ΔЕCB

ii.  ∠DBC = ∠ECB

From part (i), we already proved:

ΔDBC ≅ ΔЕCB

⇒ By CPCTC (Corresponding Parts of Congruent Triangles):

⇒ ∠DCB = ∠EBC

iii. OB = OC, where O is the point of intersection of BE and CD

Let O be the point of intersection of BE and CD.

We want to prove OB = OC.

From part (i):

• ΔDBC ≅ ΔЕCB
• So, lines CD and BE are symmetric in the isosceles triangle

Now, since these two congruent triangles are mirror images, the diagonals CD and BE intersect at a point O that lies equidistant from B and C.

So, OB = OC.