Grade 10 Maths · Euclidean Geometry

Use isosceles triangle properties

Use the fact that an isosceles triangle has two equal sides and the angles opposite them are equal, and that an equilateral triangle has all angles 60°.

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Finding Missing Angles in Isosceles Triangles | Problem Solving Using Geometric RelationshipsKyle Pearce

Angles in Isosceles Triangles // Missing Angles in Isosceles TrianglesMaths Tutor Tom

Gr 9 Geometry: Lines and TrianglesKevinmathscience

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Question 4.7 · May/June 2025 · Paper 2 · National (DBE) · 5 marks

In the diagram, M is the centre of the circle having equation

(x - 3)^{2} + y^{2} = 25

E (- 1; 3)

and T are points on the circle.

EC is a tangent to the circle at E and cuts the x-axis at D.

E D = \frac{15}{4}

units. MT is produced to meet the tangent at

C (- 7; p)

4.7 If ET is drawn, calculate the size of

E \hat{T} M

.(5)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use isosceles triangle properties

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3. Question 6.1.1 · March 2020 · Provincial · 2 marks
Proofs and Geometry problems:
6.1 JKLM is a rectangle. MLPR is a rhombus. $J \hat{M} K = R \hat{M} P = 5 5^{\circ}$ and $M \hat{R} P = 7 0^{\circ}$ . Determine:
6.1.1 $M \hat{P} R$ (2)
Show the memo (1 tick)Memo
1. ✓
  Isosceles rhombus triangle $M \hat{P} R = \frac{180 - 70}{2} = 5 5^{\circ}$ ( $∠$ s opp equal sides, $MR = RP$ )
  2 marks · Use isosceles triangle properties
4. Question 5.2.2 · March 2022 · Provincial · 3 marks
5.2 In the diagram, O is the centre of the circle and L is a point on the circumference. RLM is a tangent at L.
5.2.2 Determine with reasons the size of $\hat{L}_{2}$ .(3)
Show the memo (2 ticks)Memo
1. ✓
  Angle at centre theorem $\hat{P} = \frac{1}{2} \hat{O}_{2} = 3 2^{\circ}$ (angle at centre = 2 $\times$ angle at circ)
  1 mark · Use angle at the centre
2. ✓
  Angles opposite equal sides $\hat{L}_{2} = \hat{P} = 3 2^{\circ}$ ( $∠$ s opp equal sides, radii)
  2 marks · Use isosceles triangle properties
5. Question 4.2.2 · March 2020 · Provincial · 3 marks
4.2 In the diagram, O is the centre of the larger circle, which passes through P, R and S. R is a point on chord PV of the smaller circle PVSO. SV is produced to T. $R \hat{V} T = 9 6^{\circ} .$
4.2.2 Prove that $Δ R V S$ is isosceles.(3)
Show the memo (3 ticks)Memo
1. ✓
  Calculate S1 $\hat{S}_{1} = R \hat{V} T - \hat{R}_{2}$ (ext. $∠$ of $Δ R V S$ )
  $\hat{S}_{1} = 9 6^{\circ} - 4 8^{\circ} = 4 8^{\circ}$
  1 mark · Use exterior angle of a triangle
2. ✓
  Conclude equality $∴ \hat{S}_{1} = \hat{R}_{2} = 4 8^{\circ}$
  1 mark · Use isosceles triangle properties
3. ✓
  Final proof $∴ V S = V R$ (sides opp. equal $∠$ s)
  $Δ R V S$ is isosceles
  1 mark · Use isosceles triangle properties
6. Question 8.2 · June 2024 · Paper 2 · Provincial · 6 marks
In the figure, two circles intersect at $E$ and $D$ . $D, A, B$ and $C$ are points on the circle such that $A BC D$ is a parallelogram. $C D$ is a tangent to the smaller circle. Chords $D C$ and $BG$ intersect at $F$ . $A D G$ and $A EB$ are straight lines.
8.2 Prove, with reasons, that $D C = BG$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  S $\hat{G} = \hat{E}_{1}$
  1 mark · Use exterior angle of cyclic quadrilateral
2. ✓
  Rext $∠$ cyclic quad
  1 mark · Use exterior angle of cyclic quadrilateral
3. ✓
  S $\hat{A} = \hat{E}_{1}$ (proven in 8.1) $∴ \hat{A} = \hat{G}$ (both equal $\hat{E}_{1}$ )
  1 mark · Use isosceles triangle properties
4. ✓
  S $A B = BG$
  1 mark · Use isosceles triangle properties
5. ✓
  Rsides opp equal $∠$ s
  1 mark · Use isosceles triangle properties
6. ✓
  S & RBut $A B = C D$ (opp sides of parm) $∴ BG = C D$ (both equal $A B$ )
  1 mark · Use properties of quadrilaterals

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