Thank you for uploading the image! I'll help you solve the questions related to triangles from the provided notes. Here's a breakdown of the questions:

Question 1:

Prove for the quadrilateral PQRS:

1. $\triangle PQS \cong \triangle QPR$
2. $QS = PR$
3. $\angle PQS = \angle QPR$

Solution Steps:

• Use the property $PS = QR$ and $\angle LSPQ = \angle LRQP$ (as given).
• Utilize congruence criteria (e.g., ASA or SAS).
• Prove the congruence, equal sides, and equal angles based on geometric reasoning.

Question 2:

AB is a line segment, and P is its midpoint. Points D and E satisfy:

• $\angle BAD = \angle BAE$
• $\angle EPA = \angle LDPB$

Prove:

1. $\triangle DAP \cong \triangle EBP$
2. $AD = BE$

Solution Steps:

• Use congruence criteria such as ASA or SSS to show $\triangle DAP \cong \triangle EBP$.
• Deduce that $AD = BE$ based on corresponding parts of congruent triangles.

Question 3:

In $\triangle ABC$, AD is the perpendicular bisector of BC. Prove that $\triangle ABC$ is isosceles where $AB = AC$.

Solution Steps:

• Use the perpendicular bisector property to argue that $BD = DC$.
• Prove congruence of $\triangle ABD$ and $\triangle ACD$ using RHS criteria.
• Conclude $AB = AC$.

Question 4:

In $\triangle ABC$, altitudes $BE$ and $CF$ to $AC$ and $AB$ are equal. Prove $AB = AC$.

Solution Steps:

• Consider the property of altitudes in a triangle.
• Use the congruence of $\triangle ABE$ and $\triangle ACF$.
• Conclude that equal altitudes imply equal opposite sides.

Question 5:

For a right-angled triangle $ABC$ ($\angle A = 90^\circ$) where $AB = AC$, find $LB$ and $LC$.

Solution Steps:

• Use Pythagoras’ theorem: $BC^2 = AB^2 + AC^2$.
• Substitute $AB = AC$ to simplify and solve for $LB$ and $LC$.

Question 6:

Show that the angles of an equilateral triangle are $60^\circ$.

Solution Steps:

• Use the property of equal sides in an equilateral triangle.
• Divide $180^\circ$ equally among the three angles.

Question 7:

Prove for an isosceles triangle ($AB = AC$) with $AD$ as an altitude that:

1. $\angle B = \angle C$

Solution Steps:

• Use the property of isosceles triangles and symmetry.
• Show congruence of the two right triangles formed by the altitude $AD$.

If you'd like, I can provide detailed step-by-step solutions for any specific question from this set. Let me know which one to start with!

5 Questions Expanding on This Information:

1. How do we determine when two triangles are congruent (e.g., SAS, ASA, etc.)?
2. What is the significance of the perpendicular bisector in proving congruence?
3. Can equal altitudes in a triangle always imply equal sides? Why or why not?
4. How does the Pythagoras theorem apply differently in isosceles and right triangles?
5. Why are all angles in an equilateral triangle exactly $60^\circ$?

Tip:

Always start with the given conditions and identify congruent triangles. This often simplifies proving relationships between sides and angles.