Question

ABC is an isosceles with AB = AC .
D is the mid point of the base BC .prove that triangle ABC =triangle ADC.​

Answer 1

Answer:

SSS

Step-by-step explanation:

Given:

AB=AC

D is the mid point of BC, therefore BD=CD

Join Point D to A

In triangle ABD and triangle ACD

AB=AC

BD=CD

AD=AD (common)

Hence triangle ABD is equal to Triangle ACD by SSS

Answer 2

Given :-

• ABC is an isosceles triangle with AB = AC and D is the mid-point of base BC.

To Find:-

• Whether ΔABD ≅ ΔACD or not .

Solution :-

[ For figure refer to attachment . ]

We know that angles opposite to equal sides are equal . So here , ∠ ACD = ∠ABD

Here in ∆ADC and ∆ADB ,

• AB = AC. [ given ]
• ∠ADC = ∠ADB. [ 90° ]
• ∠ACD = ∠ABD [ just Proved ]

Therefore by AAS congruency condition , ∆ADC ≅ ∆ABD .

$\Large{\boxed{\pink{\sf Hence\:\:Proved\:!}}}$

$\rule{300}2$

Some more related Information :-

• Area of ∆ = ½ * base * height
• Area of ∆ = √[ s (s - a) (s - b) (s - c)] , where s is semiperimeter .
• Perimeter of equilateral∆ = 3*side
• Perimeter of Isosceles∆ = 2*a + b where a is equal side and c is unequal side .

$\Large{\boxed{\pink{\sf And\:we\:are\:Done!}}}$