Question

In the given figure, $\angle DAB=\angle CBA$ and AD=BC. Prove that $\angle ACD=\angle BDC$. [4 MARKS]

Answer 1

Proof: 2 Marks
Steps: 2 Marks

In $\Delta ABCand\Delta BAD$,

AD = BC [Given]

$\angle DAB=\angle CBA$ [Given]

$AB=BA$ [Common]

$\therefore \Delta ABC\cong \Delta BAD$ (By SAS congruence rule)

$\Rightarrow DB=AC$ [Corresponding parts of corresponding triangles] ...... (1)

In $\Delta ADCand\Delta BCD$

$AD=BC$ [Given]

$DB=AC$ [From (1)]

$DC=CD\left(Common\right)$

$\Delta ADC\cong \Delta BCD$ [By SSS congruence rule]

$\Rightarrow \angle ACD=\angle BDC$ [Corresponding parts of corresponding triangles]