Question

in an isoscles triangle ABC , with AB= AC ,D is the mid point of side BC of triangle ABC show that triangle ABC congurent triangle ACD​

Answer 1

Refer to the attachment.

Before we solve

We have to prove the congruence of the two triangles. We will use one between three similarity criteria, which are SSS, SAS, ASA.

Solution

First, D will divide $\overline{BC}$ into equal lengths.

$\rightarrow\overline{BD}=\overline{CD}$ ...[I]

Two triangles share $\overline{AD}$ ...[II]

Given that $\overline{AB}=\overline{AC}$ ...[III]

By SSS criteria, two triangles are congruent.

Hence $\triangle{ABD}\cong\triangle{ACD}$.

Answer 2

$\Large{\textsf{\textbf{\underline{\underline{Given\::}}}}}$

As shown in the figure,

In an isosceles ∆ ABC,

• AB = AC

• D is the midpoint of BC.

$\Large{\textsf{\textbf{\underline{\underline{To\:do\::}}}}}$

• $\rm{\triangle{ABC}\:\cong\:\triangle{ACD}\:}$

$\Large{\textsf{\textbf{\underline{\underline{Solution\::}}}}}$

In ∆ABC & ∆ACD,

⠀❶ AB = AC

⠀❷ AD = AD [ $\because$ AD is a common line.]

⠀❸ BD = CD [ $\because$ D is the midpoint of BC.]

From the above solution,

$\bf\pink{\triangle{ABC}\:\cong\:\triangle{ACD}\:}$ [S-S-S criterion]

⠀⠀⠀⠀⠀$\red\star\:\:{\textsf{\textbf{\underline{\blue{Hence\:proved}}}}}\:.\:\red\star$