Question

answer the questions

Answer 1

By RHs property the triangle is congruent

Answer 2

$\bold{\large{Solution\: \colon}}$

$\bold{\large{\underline{Given \: \colon}}}$

∆ ABC in which AD is perpendicular on BC and AB = AC

$\bold{\large{\underline{To \: Find}}}$

$\bold{ \triangle{ADB} \: \cong \: \triangle{ADC}}$

$\bold{\large{\underline{Solution\: \colon}}}$

In triangles

$\bold{ \triangle{ADB} \: \& \: \triangle{ADC}}$

$\bold{ AD \: = \: AD}$ ....(common)

$\bold{AB \: = \: AC }$ ....(given)

$\bold{ \angle{ADB}\: = \: \angle{ADC}}$ ....(90°)

Now it is clear that

$\bold{ \triangle{ADB} \: \cong \: \triangle{ADC}}$

By $\bold{RHS}$ Congruency Rule

AB = AC (given)

by this

$\angle{ABD} \: = \: \angle{ACD}$

If the $\bold{opposite\: sides}$ of triangle are equal then it's $\bold{opposite}$ angles are also equal.

$\bold{ \triangle{ADB} \: \& \: \triangle{ADC}}$

$\bold{ \angle{ADB}\: = \: \angle{ADC}}$ ....(90°)

$\bold{\angle{ABD} \: = \: \angle{ACD}}$ ....(proved)

$\bold{AB \: = \: AC }$ ....(given)

$\bold{ \triangle{ADB} \: \cong \: \triangle{ADC}}$

By $\bold{AAS}$ Congruency rule

$\bold{\large{\underline{Answer\: \colon}}}$

$\bold{\large{\boxed{\boxed{(4)\: Both \: (2) \: \& \: (3)}}}}$

Hence Proved

$\bold{\large{Thanks}}$