Question

O is centre of circle seg AB is
diameter seg OA=seg AP points
O,A&P are collinear.
line PC is tangent which touches
at pt C . the tangent passing
through pt A intersect the line
PC at pt E &line BC at pt D
prove that triangle CED is an equilateral triangle.
( construction - draw seg OC) ​

Answer 1

Triangle CED is an equilateral triangle.

Step-by-step explanation:

Given data

Centre of the circle  - O

Diameter of the circle - AB

OA = AP

A and P are collinear

Prove that triangle CED is an equilateral triangle

Draw the segment OC

OCP is a right angle triangle, where PC is perpendicular to OC

Radius of the segment OC is r

OC = r

OP = OA + AP

Where AP = OA

OP = OA + OA

Substitute OA = r in above relation

OP = r + r

OP = 2r

$\text{PC} = \sqrt{\text{OP}^2 - \text{OC}^2} = \sqrt{4 \text{r}^2 - \text{r}^2} = \sqrt{ 3 \text{r}^2} = \sqrt{3} \ (\text{r} )$

In triangle OCP, the ratio of the sides is,

$\text{OC} : \text{CP} : \text{PO} = \text{r} : \text{r} \sqrt{3} : 2\text{r}$

$\text{OC} : \text{CP} : \text{PO} = 1 : \sqrt{3} : 2$

It is cleared that the angles should be 30°, 60° and 90°

Then, ∠OPC = 30° and ∠COP = 60° ---------> (1)

So ,  ∠APE = 30°.

In Triangle APE, ∠PAE = 90°

∠APE = 30°, ∠AEP = 60° and ∠CED = 60° ( Vertically opposite angles)

From the equation (1)  (∠COP = 60°)

∠COB = 180° -60° = 120°

In the triangle BOC ,  OC = OB

∠OCB = ∠OBC

          $\Rightarrow \frac{1}{2} (180^\circ - 120^\circ) = 30^\circ$

So, in triangle PBC

∠P = ∠B which is equal to 30°

∠PCD = 30° + 30° = 60°

The exterior angles of a triangle is equal to the sum of the interior opposite angles.  

Therefore, CED is an equilateral triangle

To Learn More ...

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