Prove that: (i) ABD ACD (ii) B= C (iii) ADB = ADC (iv) A D B=90^ - Vidyadip

Question

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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