Question

LMN is an isosceles triangle with side LM = LN.
A, B, and C are the midpoints of LM, MN, and NL,
respectively. What is the congruence condition that
proves CAM ~=ACN?

also please send a image of triangle with all measurements​

Answer 1

$\huge \dag \underline{ \boxed{ \mathfrak{ \purple{ answer}}}}$

 

Given: Δ LMN is an isosceles triangle, m∠LMN = m∠LNM

And, LP bisects angle NLQ.

Prove: LP ║ MN

Now, In triangle LMN,

m∠LMN + m∠LNM+ m∠MLN = 180° (By the property of triangle)

⇒ m∠LMN + m∠LMN + m∠MLN = 180° ( Here,  m∠M = m∠N )

⇒ 2 m∠LMN + m∠MLN = 180° -------(1)

Now, LP bisects angle NLQ.

⇒ m∠PLN =m∠QLP ( by the property of angle bisector)

Since, m∠QLP + m∠PLN + m∠MLN = 180° ( sum of all angles on a straight line)

m∠QLP + m∠QLP + m∠MLN = 180°

2 m∠QLP + m∠MLN = 180°

⇒ m∠MLN = 180°- 2 m∠QLP --------(2)

From equation (1) and (2),

2 m∠LMN + 180°- 2 m∠QLP = 180°

2 m∠LMN - 2 m∠QLP = 0

2 m∠LMN = 2 m∠QLP

m∠LMN = m∠QLP

⇒ ∠LMN ≅ ∠QLP

Thus, By the inverse of corresponding angle theorem,

LP ║ MN

Answer 2

$\large \green{ \fcolorbox{gray}{black}{ ☑ \: \textbf{Verified \: answer}}}$

By the inverse of corresponding angle theorem,

LP ║ MN