Question

28. ABC and DBC are two isosceles triangles on the same base BC.
Show that
(i triangleABD is equal to triangleACD
(ii) AD bisects angleA​

Answer 1

Answer:

Given

ΔABC and ΔDBC are two isosceles triangles in which AB=AC & BD=DC.

To Prove:

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Proof:

(i) In ΔABD and ΔACD,

AD = AD (Common)

AB = AC (given) .

BD = CD (given)

Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)

∠BAD = ∠CAD (CPCT)

∠BAP = ∠CAP

(ii) In ΔABP & ΔACP,

AP = AP (Common)

∠BAP = ∠CAP

(Proved above)

AB = AC (given)

Therefore,

ΔABP ≅ ΔACP

(by SAS congruence rule).

(iii) ∠BAD = ∠CAD (proved in part i)

Hence, AP bisects ∠A.

also,

In ΔBPD and ΔCPD

PD = PD (Common)

BD = CD (given)

BP = CP (ΔABP ≅ ΔACP so by CPCT)

Therefore, ΔBPD ≅ ΔCPD (by SSS congruence rule)

Thus,

∠BDP = ∠CDP( by CPCT)

Hence, we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD

(by CPCT as ΔBPD ≅ ΔCPD)

& BP = CP (CPCT)

∠BPD + ∠CPD = 180° (BC is a straight line)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90°

Hence, AP is the perpendicular bisector of BC.

Thanks

Have a colossal day ahead

Answer 2

Step-by-step explanation:

Solution :

• Answer i)

To Prove : ∆ ABD = ∆ ACD

In ∆ ABD and ∆ ACD -

AB = AC ------ (Given)

BD = CD ------ (Given)

AD = AD ------ (Common arm)

Therefore, ∆ ABD ≅ ∆ ACD by SSS congrueny.

_______________________________

• Answer ii)

To Prove : AD bisects ∠A

∠A = ∠CAD + ∠BAD

As proven above, ∆ ABD ≅ ∆ ACD,

∠CAD = ∠BAD ------ (By C.P.CT)

Halves of ∠A are equal.

Therefore, AD bisects ∠A.