Question

AD is an altitude of an isosceles triangle ABC in which AB = AC . Which congruence condition will you use to prove ∆ADB congruent to ∆ ADC ? {Class 9}

Answer 1

Given :-

AD is an altitude of an isosceles triangle ABC in which AB = AC.

Required to find :-

• which congruency condition will be used to prove ∆ADB congruent to ∆ ADC ?

Solution :-

Diagram :-

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Given data :-

AD is an altitude of an isosceles triangle ABC in which AB = AC.

we need to find the which congruency condition used to prove that ∆ ABD is congruent to ∆ ADC .

So,

In order to find the congruency condition first find whether ∆ ABD is Congruent to ∆ ADC .

Here,

Consider ∆ ABC in which AD is an altitude , AB = AC .

In ∆ ABC , AD is an altitude which divides the triangle into 2 triangle namely ∆ ADB & ∆ ADC

So,

In ∆ ADB & ∆ ADC

AD = AD ( side )

[ Reason :- common side ]

∠ADB = ∠ADC ( angle )

[ Reason :- Right angle ( 90° ) ]

AB = AC ( side )

[ Reason :- Given Information ]

Hence,

By using S.A.S. congruency rule

$\tt \Delta ADB \cong \Delta ADC$

Therefore,

The congruency condition which we used to prove that ∆ ADB is congruent to ∆ ADC = S.A.S. rule/ criteria

Additional Information :-

In the above question , we can also use S.S.S. Congruency rule to prove that ∆ ADB is congruent to ∆ ADC .

In a triangle , an altitude will be perpendicular to the side which is drawn from the opposite vertex .

Since, the altitude falls on the base perpendicularly the side will be divided into 2 equal halves .

In a triangle , we can draw 3 Altitudes .

Some of other congruency rules are ;

• SAS ( Side , angle , side )

• AAA ( angle , angle , angle )

• SSS ( side , side , side )

• RHS ( Right angle , Hypotenuse , Side )

Answer 2

Given :

• In triangle ΔABC , AD is an altitude
• AB = AC

To find :

$\sf ΔADB \cong ΔADC$

Solution :

Now , in triangle ΔADB And ΔADC

$\sf \mapsto AD = AD \: \: \: (\because \: common \: side)</p><p> \\ \\ \sf \mapsto \angle ADB = \angle ADC \: \: \: (\because \: angle = 90) </p><p> \\ \\ \sf \mapsto</p><p></p><p>AB = AC \: \: \: ( \because \: given)$

$\sf \therefore \underline{ΔADB \cong ΔADC \: by \:SAS \: congruence \: rule }$