Prove the theorem that states XDR = DYX

Question

ScanSolve AI Tutor · Accepted Answer

Here is the proof for the given theorem.

1.1 Prove the theorem that states $\angle XDR = \angle DYX$

Construction.
Draw a diameter through the point of tangency D. Let this diameter be DP, where P is a point on the circumference. Join PX.
Angle between tangent and diameter.
The tangent RS is perpendicular to the diameter DP at the point of contact D.
$\angle RDP = 90^\circ$
Angle in a semicircle.
The angle subtended by a diameter at any point on the circumference is $90^\circ$ .
$\angle DXP = 90^\circ$
Angles in a triangle.
Consider $\triangle DXP$ . The sum of angles in a triangle is $180^\circ$ .
$\angle XDP + \angle DPX + \angle DXP = 180^\circ$
Substitute $\angle DXP = 90^\circ$ :
$\angle XDP + \angle DPX + 90^\circ = 180^\circ$ $\angle XDP + \angle DPX = 90^\circ \quad (1)$
Relate angles at the point of tangency.
From Step 2, we know $\angle RDP = 90^\circ$ . We can express $\angle RDP$ as the sum of $\angle RDX$ and $\angle XDP$ .
$\angle RDX + \angle XDP = 90^\circ \quad (2)$
Equate expressions for $90^\circ$ .
From equations (1) and (2):
$\angle XDP + \angle DPX = \angle RDX + \angle XDP$
Subtract $\angle XDP$ from both sides:
$\angle DPX = \angle RDX$
Angles in the same segment.
Angles subtended by the same arc at the circumference are equal. Both $\angle DPX$ and $\angle DYX$ are subtended by the arc DX.
$\angle DPX = \angle DYX$
Conclusion.
From Step 6 and Step 7, we can conclude:
$\boxed{\text{}\angle XDR = \angle DYX\text{}}$
This proves the Tangent-Chord Theorem.

Answer