Question

Identify the missing parts in the proof. Given: ∠ABC is a right angle. DB bisects ∠ABC. Prove: m∠CBD = 45°

Answer 1

Answer with Step-by-step explanation:

We are given that

Angle ABC is  a right angle .

$\angle ABC=90^{\circ}$

DB bisects the angle ABC.

We have to prove that $m\angle CBD=45^{\circ}$

When DB bisects the angle ABC

$m\angle ABD=m\angle CBD$

$m\angle CBD+m\angle ABD=\angle ABC$

Substitute the value

$m\angle CBD+m\angle CBD=90$

$2m\angle CBD=90$

$m\angle CBD=\frac{1}{2}\times 90=45^{\circ}$

Hence, proved.

Answer 2

When an angle is bisected, then it would be divided into two equal parts. Thus, bisection of <B implies m∠CBD = 45°.

Bisection of angle implies that the angle would be divided into two equal parts.

Given that ΔABC is a right angled triangle.  Thus since <ABC is right angle, it implies that the measure of the angle is $90^{o}$.

Then;

DB is not perpendicular to AC

m<A + m<ABD + m<ADB = $180^{o}$ (sum of angles in a triangle)

Also,

m<C + m<CBD + m<CDB = $180^{o}$ (sum of angles in a triangle)

Since DB is a bisector of right angle B, then;

m<B = m<ABD + m<CBD (bisection property of an angle)

$90^{o}$ = m<ABD + m<CBD

But,

m<ABD ≅ m<CBD (similarity property)

So that:

$90^{o}$ = 2<CBD

m<CBD = $\frac{90^{o} }{2}$

          = $45^{o}$

m<CBD = $45^{o}$

Thus, the measure of angle CBD = ABD = $45^{o}$

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