[5 marks sum]

Question 15 Marks

Prove that:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\angle B=\angle C$
(iii) $\angle ADB =\angle ADC$
(iv) $\angle A D B=90^{\circ}$

Answer

Given: In the figure,
$
\begin{aligned}
& A D=A C \\
& B D=C D
\end{aligned}
$

To prove:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\angle B=\angle C$
(iii) $\angle ADB =\angle ADC$
(iv) $\angle ADB =90^{\circ}$
Proof: In $\triangle A B D$ and $\triangle A C D$
$A D=A D$ .................(common)
$A D=A C$ .................(given)
$B D=C D$ ................(given)
(i) $\therefore \triangle ABD \cong \triangle ACD$ ..............(SSS axiom)
(ii) $\therefore \angle B=\angle C$ (c.p.c.t.)
(iii) $\angle ADB =\angle ADC$ (c.p.c.t.)
But $\angle ADB +\angle ADC =180^{\circ}$ ............(Linear pair)
$\begin{aligned} & \therefore \angle ADB =\angle ADC \\ & \text { (iv) } \angle ADB =\angle ADC \\ & =\frac{180^{\circ}}{2} \\ & =90^{\circ}\end{aligned}$

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Question 25 Marks

Prove that:
(i) $\triangle A B C \cong \triangle A D C$
(ii) $\angle B=\angle D$
(iii) $A C$ bisects angle $D C B$

Answer

Given: In the figure,
$
A B=A D, C B=C D
$
To prove: $\triangle A B C \cong \triangle A D C$
$
\angle B=\angle D
$
AC bisects angle DCB

Proof: In $\triangle A B C$ and $\triangle A D C$,
$A C=A C$ ..............(common)
$A B=A D$ ..............(given)
$C B=C D$ ..............(given)
(i) $\therefore \triangle ABC \cong \triangle ADC$ .................(SSS aciom)
(ii) $\therefore \angle B=D$ .................(c.p.c.t.)
$
\angle B C A=\angle D C A
$
(iii) $\therefore A C$ bisects $\angle D C B$

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MATHS [5 marks sum]

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