Question

1. in the figure shown, $\angle nwl \cong \angle hwl$ and $lw \perp nh$. prove $\triangle nlw \cong \triangle hlw$

Explanation:

Step1: Identify right angles

Since \( LW \perp NH \), \( \angle LWN = \angle LWH = 90^\circ \).

Step2: Identify common side

\( LW \) is a common side to both \( \triangle NLW \) and \( \triangle HLW \), so \( LW = LW \) (reflexive property).

Step3: Identify congruent angles

Given \( \angle NWL \cong \angle HWL \). Also, from the diagram, \( \angle N = \angle H = 28^\circ \) (or we can use the given angle congruence \( \angle NWL \cong \angle HWL \) and the right angles).

Step4: Apply ASA congruence

In \( \triangle NLW \) and \( \triangle HLW \):

• \( \angle NWL = \angle HWL \) (given)
• \( LW = LW \) (common side)
• \( \angle LWN = \angle LWH = 90^\circ \) (from perpendicularity)

By the Angle - Side - Angle (ASA) congruence criterion, \( \triangle NLW \cong \triangle HLW \).

Answer:

To prove \( \triangle NLW \cong \triangle HLW \):

1. Right Angles: Since \( LW \perp NH \), \( \angle LWN=\angle LWH = 90^\circ \) (definition of perpendicular lines).
2. Common Side: \( LW \) is a side of both \( \triangle NLW \) and \( \triangle HLW \), so \( LW = LW \) (reflexive property of congruence).
3. Given Angle Congruence: \( \angle NWL\cong\angle HWL \) (given).

By the Angle - Side - Angle (ASA) congruence postulate (two angles and the included side of one triangle are congruent to two angles and the included side of the other triangle), we conclude \( \triangle NLW\cong\triangle HLW \).