Question

ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

Answer 1

ΔABC is an isosceles triangle     (given)

ΔDBC is an isosceles triangle      (given)

Both ΔABC and ΔDBC lies on same base BC     (given)

 

To be proof :

$\angle{ABD}$ = $\angle{ACD}$

 

So,

$\angle{ABC}$ = $\angle{ACB}$   [ as ΔABC is an isosceles triangle ]

and

$\angle{DBC}$ = $\angle{DCB}$   [ as ΔDBC is an isosceles triangle ]

 

construction :

Join A to D

 

now,

In ΔABD and ΔACD

AB = AC     (given)

BD = CD     (given)

AD = AD     (common side)  

Therefore,

by SSS(Side-Side-Side) congruence condition ΔABD ≅ ΔACD

 

Hence,

$\angle{ABD}$ = $\angle{ACD}$  by CPCT