1. In the equilateral triangle ABC, the base BC is extended to point E in such a way that CE = BC. By joining A and E, find the circular values (in radians) of the angles of triangle ACE.

2. Given: In triangle △ABC, O is the circumcenter and OD ⊥ BC. Prove that: ∠BOD = ∠BAC Let’s break it down in English: **Given:** In triangle △ABC, O is the circumcenter (the point where the perpendicular bisectors of the sides meet), and OD is perpendicular to side BC. **To Prove:** The angle ∠BOD formed at the center between points B and D is equal to the angle ∠BAC at the vertex A. This is a classic geometry result based on the properties of a circle and triangle. Would you like me to walk you through the full proof in English as well?

3. Draw an equilateral triangle ABC with each side measuring 5 cm. Then, draw the circumcircle of that triangle. At point A on the circle, draw a tangent. On the tangent, take a point P such that AP = 5 cm. From point P, draw another tangent to the circle, and write down which point on the circle this second tangent touches.

4. Let the smaller pillar be AB \(= x\) meters and the larger pillar be CD \(= 2x\) meters.
The midpoint of the base connection BD is O; the angles of elevation from O to the tops of the pillars are \(\angle\)AOB = \(\theta\) and \(\angle\)COD = 90\(^o-\theta\).
Since O is the midpoint of BD, we have BO = OD = \(\frac{120}{2}\) meters = 60 meters.

From \( \triangle \)ABO, we get:
\(\cfrac{AB}{BO} = \tan \theta\)
Or, \(\cfrac{x}{60} = \tan\theta\) --------(i)

From \( \triangle \)COD, we get:
\(\cfrac{CD}{OD} = \tan(90^o - \theta)\)
Or, \(\cfrac{2x}{60} = \cot\theta\) --------(ii)

Multiplying equations (i) and (ii), we get:
\(\cfrac{x}{60} \times \cfrac{2x}{60} = \tan\theta \times \cot\theta\)
Or, \(\cfrac{2x^2}{60 \times 60} = 1\)
Or, \(x^2 = \cfrac{60 \times \cancel{60}30}{\cancel{2}}\)
Or, \(x = 30\sqrt2\)

\(\therefore\) The height of the smaller pillar is \(30\sqrt2\) meters.
And the height of the larger pillar is \(30\sqrt2 \times 2\) meters \(= 60\sqrt2\) meters.
(Proved).

5. From vertex A of triangle \(\triangle ABC\), a perpendicular AD is drawn to the base BC, intersecting BC at point D. If \(AD^2 = BD \cdot CD\), prove that triangle ABC is a right-angled triangle and that \(\angle A = 90^\circ\).

6. "Chords PQ and RS of a circle intersect each other at point X inside the circle. By joining P to S and R to Q, prove that triangles ∆PXS and ∆RSQ are similar. From this, prove that: PX × XQ = RX × XS Or, when two chords of a circle intersect internally, the product of the two segments of one chord is equal to the product of the two segments of the other chord.