Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D
(iii) AC bisects angle DCB.

Open in App

Solution

(i) In the given figure
AB = AD and CB = CD
In ∆ ABC and ∆ ADC
AC = AC is common
It is given that
AB = AD and CB = CD
So, ∆ ABC ≅ ∆ ADC (SSS Axiom)

(ii) ∠B = ∠D (By C.P.C.T)

(iii) So, we get
∠BCA = ∠DCA (by C.P.C.T)
Therefore, AC bisects ∠DCB.

Suggest Corrections

Similar questions

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that (i) AC bisects $∠ A a n d ∠ C$ , (ii) BE = DE, (iii) $∠ A B C = ∠ A D C$ .

Q. (a) In the given figure, prove that:

(i)

Δ A B C ≅ Δ D C B

(ii) AC = DB

(b) In the figure, it is given that

∠

A =

90^{\circ}

, AB = AC, and D is the mid point of BC. Find

∠

ADC.

[4 MARKS]

In the given figure, AB = AC. Prove that :

(i) DP= DQ

(ii) AP = AQ

(iii) AD bisects angle A

$In the given figure, A B | | E C, A B = A C a n d A E b i s e c t s ∠ DAC.Prove that:$

$(i) ∠ E A C = ∠ A C B (i i) ABCE is a parallelogram.$

In a triangle ABC, AC = AB and the altitude AD bisects BC. Prove that ΔADC ≅ ΔADB.

Join BYJU'S Learning Program

Prove that: (i) ∆ ABC ≅ ∆ ADC (ii) ∠B = ∠D (iii) AC bisects angle DCB.

(i) In the given figure AB = AD and CB = CD In ∆ ABC and ∆ ADC AC = AC is common It is given that AB = AD and CB = CD So, ∆ ABC ≅ ∆ ADC (SSS Axiom) (ii) ∠B = ∠D (By C.P.C.T) (iii) So, we get ∠BCA = ∠DCA (by C.P.C.T) Therefore, AC bisects ∠DCB.

Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D
(iii) AC bisects angle DCB.

(i) In the given figure
AB = AD and CB = CD
In ∆ ABC and ∆ ADC
AC = AC is common
It is given that
AB = AD and CB = CD
So, ∆ ABC ≅ ∆ ADC (SSS Axiom)

(ii) ∠B = ∠D (By C.P.C.T)

(iii) So, we get
∠BCA = ∠DCA (by C.P.C.T)
Therefore, AC bisects ∠DCB.