Question

prove that:
(i) triangle ABC = triangle ADC
(ii) angle B = angle D
(iii) AC bisects angle DCB​

Answer 1

\: \: \: ༄༄Answer༄༄

$\huge {\underline{ Given}:-}$

ABCD is a quadrilater

side DC = CB

and DA = AB

$\huge {\underline{ To\: prove }:-}$

(i) triangle ABC ≅ triangle ADC

(ii) angle B = angle D

(iii) AC bisects angle DCB

$\huge {\underline{ Proof}:-}$

$\: \: \: In \: triangle\: ∆ ABC\: and\: ∆ ADC$

DC = CB [Given]

DA = AC [Given]

AC = AC [Common]

∆ ABC ≅ ∆ ADC [by SSS rule]

angle B = angle D [by CPCT ]

∆ ABC ≅ ∆ ADC Hence, AC bisects angle DCB

$\huge {\underline{\boxed{Hence Proved}}}$

Answer 2

Answer:

Consider triangle ABC and triangle ACD

AB=AD (Given in the figure)

BC=CD (Given in the figure)

AC=AC (Common side/identity)

Triangle ABC=Triangle ADC (SSS Condition)

Angle B= Angle D (CPCTC)

∆ ABC ≅ ∆ ADC Hence, AC bisects angle DCB

Hence proved

Step-by-step explanation:

SSS- SIDE SIDE SIDE Condition

CPCTC- Corresponding parts of congruent triangles are congruent

Hope this answer is helpful for you. Mark me brainlist and thank my answer