ABCD is a trapezium in which BC ∥ AD and AC = CD. If ∠ABC = 69° and ∠BAC = 23°, then what is the measure of ∠ACD (in degree)?​

Correct option is A

Given:

ABCD is a trapezium (trapezoid) with BC parallel to AD (BC ∥ AD).
AC = CD (This means triangle ACD is an isosceles triangle).
Angle ABC (∠ABC) = 69°.
Angle BAC (∠BAC) = 23°.
Find: The measure of Angle ACD (∠ACD).

Solution:
Find Angle ACB in Triangle ABC.

The sum of angles in any triangle is 180°.

In Triangle ABC:
∠ACB = 180° − (∠ABC + ∠BAC)
∠ACB = 180° − (69° + 23°)
∠ACB = 180° − 92°
∠ACB = 88°.

Use the property of parallel lines to find Angle CAD.

Since BC is parallel to AD (BC ∥ AD) and AC is a transversal line, the alternate interior angles are equal.

∠CAD = ∠ACB

Since ∠ACB = 88° (from Step 1), then ∠CAD = 88°.

Find Angle ACD in Triangle ACD.

We are given that AC = CD. This means Triangle ACD is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal.

The angle opposite side CD is ∠CAD.
The angle opposite side AC is ∠CDA.

Therefore, ∠CDA = ∠CAD = 88°.

Now, apply the sum of angles property to Triangle ACD:

∠ACD + ∠CAD + ∠CDA = 180°
∠ACD + 88° + 88° = 180°
∠ACD + 176° = 180°

∠ACD = 180° − 176°

∠ACD = 4°

The measure of ∠ACD is 4 degrees.