Question

The points A, B, C and D lie on a circle. CDE is a straight line. BA=BD, CB=CD, Angle ABD =32 degrees Work out the size of angle ADE. You must give a reason for each stage of working.

Answer 1

Answer:

$\angle ADE =$ $69^\circ$

Step-by-step explanation:

Given:

$\angle ABD = 32^\circ$

BA = BD

CB = CD

In $\triangle ABD:$

BA = BD

We know that in a triangle, angles opposite to equal sides will also be equal.

$\Rightarrow \angle BAD = \angle BDA$

We know that sum of internal angles of a triangle is equal to $180^\circ$

$\angle BAD + \angle BDA +\angle ABD=180^\circ\\\Rightarrow \angle BAD + \angle BDA +32=180\\\Rightarrow \angle BAD + \angle BDA = 180 -32\\\Rightarrow \angle BAD + \angle BDA = 148\\\Rightarrow \angle BAD = \angle BDA = \dfrac{148}{2} = 74^\circ$

Quadrilateral ABCD is cyclic so, sum of opposite angles is equal to $180^\circ$.

$\angle BAD + \angle BCD = 180^\circ\\\Rightarrow 74 + \angle BCD = 180\\\Rightarrow \angle BCD = 180-74 = 106^\circ$

In $\triangle BCD:$

BC = CD

We know that in a triangle, angles opposite to equal sides will also be equal.

$\Rightarrow \angle DBC = \angle CDB$

We know that sum of internal angles of a triangle is equal to $180^\circ$

$\angle DBC + \angle BCD +\angle CDB=180^\circ\\\Rightarrow \angle DBC + 106 +\angle CDB=180\\\Rightarrow \angle DBC +\angle CDB=180 - 106\\\Rightarrow \angle DBC =\angle CDB=\dfrac{74}{2} = 37^\circ$

It is given that the line CDE is a straight line, so the sum of angles on one side of point D is $180^\circ$.

$\angle CDB + \angle BDA +\angle ADE = 180^\circ\\\text{Putting values of } \angle CDB \text{ and }\angle BDA:\\\Rightarrow 37 + 74 + \angle ADE = 180\\\Rightarrow \angle ADE = 180 -111\\\Rightarrow \angle ADE = 69^\circ$

So, the answer is $\angle ADE =$ $69^\circ$

Answer 2

The size of ∠ADE of the cyclic quadrilateral is;

∠ADE = 53°

We are given;

BA = BD

CB = CD

∠ABD = 32°

Now, ΔABD is an isosceles triangle and since sum of angles in a triangle is 180°, then;

∠ABD + ∠BAD + ∠BDA = 180°

∠BAD + ∠BDA = 180 - 32

∠BAD + ∠BDA = 148°

Since isosceles, then;

∠BAD = ∠BDA = 148/2

∠BAD = ∠BDA = 74°

Now, in cyclic quadrilaterals, opposite angles are supplementary and as such add up to 180°. Quadrilateral ABCD is cyclic and Thus;

∠BAD + ∠BCD = 180°

74 + ∠BCD = 180°

∠BCD = 180 - 74

∠BCD = 106°

ΔBCD is an isosceles triangle and as such;

∠CBD = ∠CDB = 106/2

∠CBD = ∠CDB = 53°

Now, sum of angles on a straight line add up to 180°. Thus;

∠CBD + ∠BDA + ∠ADE = 180°

Plugging in the relevant values gives;

53° + 74° + ∠ADE = 180°

∠ADE = 180 - (53 + 74)

∠ADE = 53°

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