Question

5. ZXYZ is bisected by YP. L is any point on YP and MLN is perpendicular to YP. Prove that LM=LN.​

Answer 1

$\fbox{Answer }$

Given:

➪.∠XYZ is bisected by YP which means ∠XYP=∠ZYP, L is any point on YP and MLN is perpendicular to YP that is ∠YLM=∠YLN=90°.

To prove:

➪.LM=LN

Proof:

➪. In ΔMYL and ΔNYL

We have,

• ∠YLM=∠YLN=90°( MLN is perpendicular to YP)

• YL=YL( common)

• ∠MYL=∠NYL(Given)

Therefore, by ASA rule of congruency,

$\boxed{ΔMYL ≅ ΔNYL}$

⇒By CPCTC, LM=LN

Hence proved.

Answer 2

$\huge \mathscr{\orange {\underline{\red{\underline {Answer:-}}}}}$

GIVEN :-

■ ZXYZ is bisected by YP.

■ L is any point on YP and MLN is perpendicular to YP.

TO PROVE :-

Prove that LM = LN.

PROOF :-

In triangle YML and triangle YNL,

∠MYL = ∠NYL < ∠XYZ is bisected by YP

YL = YL (common)

∠MLY = ∠NLY ( both are 90° )

By ASA Criterion Rule,

              △YML ≅△ YNL.

Also,

By CPCT,

        LM = LN

• HENCE PROVED •
• I HOPE IT'S HELPFUL.