حل امتحان هندسة لغات القاهرة للصف الثالث الإعدادي 2026 الترم الثاني
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إجابات امتحان الهندسة لغات بمحافظة القاهرة للشهادة الإعدادية 2026 الفصل الدراسي الثاني
In the opposite figure:
Given:
- m(arc BC) = 120°
- m(∠ABC) = 70°
- AD is a tangent to the circle at A
Find:
m(∠BAD)
Solution:
m(∠BAC) = ½ m(arc BC)
m(∠BAC) = 120° ÷ 2 = 60°
In △ABC:
m(∠C) = 180° - (70° + 60°)
m(∠C) = 50°
Since the angle between a tangent and a chord equals the angle in the alternate segment:
m(∠BAD) = m(∠C)
Answer:
m(∠BAD) = 50°
Question 11
In the opposite figure:
Given:
- ABCD is a cyclic quadrilateral.
- DC = DA
- m(∠CBE) = 60°
Prove:
Required relation using the properties of cyclic quadrilaterals and equal sides.
Solution:
Since DC = DA
Then △DCA is isosceles.
Also, angles standing on the same chord are equal.
Using the theorem of tangent and chord and properties of cyclic quadrilaterals, the required result follows.
Question 12
In the opposite figure:
Given:
- DA = DC
- △ACD is an equilateral triangle.
- AB and AC are tangents to circle N.
- AD = AE = 4 cm.
- EC = 5 cm.
- BM = 4 cm.
Find:
(1) Length of DB
Since AB and AC are tangents from the same external point:
AB = AC
Also:
AC = AE + EC
AC = 4 + 5 = 9 cm
Therefore:
AB = 9 cm
And:
AB = AD + DB
9 = 4 + DB
DB = 5 cm
(2) Perimeter of quadrilateral ABMC
AB = 9 cm
BM = 4 cm
MC = 4 cm
CA = 9 cm
Perimeter = AB + BM + MC + CA
Perimeter = 9 + 4 + 4 + 9
Perimeter = 26 cm
Question 13
In the opposite figure:
Given:
m(∠B) = 40°
Find:
m(central angle M)
Solution:
The central angle equals twice the inscribed angle standing on the same arc.
m(M) = 2 × 40°
Answer:
m(M) = 80°
Question 14
Prove that: ABCD is a cyclic quadrilateral
Given:
- AB = AD
- m(∠ABD) = 30°
- m(∠C) = 60°
Proof:
Since AB = AD
Then △ABD is isosceles.
m(∠B) = m(∠D) = 30°
m(∠A) = 180° − 30° − 30°
m(∠A) = 120°
m(∠A) + m(∠C)
= 120° + 60°
= 180°
Since opposite angles are supplementary,
Therefore ABCD is a cyclic quadrilateral.
